At school, I really struggled to understand the concept of imaginary numbers. My teacher told us that an imaginary number is a number that has something to do with the square root of $-1$. When I tried to calculate the square root of $-1$ on my calculator, it gave me an error. To this day I still do not understand imaginary numbers. It makes no sense to me at all. Is there someone here who totally gets it and can explain it?

Why is the concept even useful?

Simon Fraser
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Sachin Kainth
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    This isn't exactly the same question, but you might find some of the answers helpful: http://math.stackexchange.com/questions/154/do-complex-numbers-really-exist. Also, have you looked at [the Wikipedia article](http://en.wikipedia.org/wiki/Imaginary_number)? It's easier to give an answer that's useful to you if you point out what you find missing or unclear there. – joriki Sep 20 '12 at 12:23
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    Consider first the real numbers. Each of them can be imagined as a point on a line (or ruler if you like). Each point is a number and vice versa. In comparison the imaginary numbers can be viewed as points on a plane. Each point is a complex(imaginary) number. – ivan Sep 20 '12 at 12:25
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    I don't get it. – Sachin Kainth Sep 20 '12 at 12:28
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    I thought i = squareroot(-1) – Sachin Kainth Sep 20 '12 at 12:29
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    @SachinKainth: What is a real number? I mean, what do you understand a real number to be and why do you not struggle with that concept? If people see that you understand such numbers for particular reasons, they may be able to give similar reasons for the existence of complex numbers, or at least gauge what would be required to convince you that complex numbers are useful. – Michael Albanese Sep 20 '12 at 12:54
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    See also [Do complex numbers really exist?](http://math.stackexchange.com/a/2658/242) – Bill Dubuque Sep 20 '12 at 13:10
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    $\mathbb{R}eally$ exist? – user1729 Sep 20 '12 at 13:14
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    $i$ means rotating $90^{\circ}$ – chaohuang Sep 20 '12 at 14:58
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    @SachinKainth Hey Sachin, don't worry about not getting it too much. If I didn't understand it and read any of the answers on this page, I wouldn't get it either. I recommend reading from as many sources as you can and try to find one that explains it to you in a way you understand. Just wanted to say, don't feel bad about it. Years ago, it took me a while to get it, until I found a perfect explanation somewhere. – Korgan Rivera Sep 20 '12 at 15:06
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    @ivan Your comment is misleading. A *complex* number is a number on the plane. An *imaginary* number is merely the second coordinate in 2D, the imaginary part of the complex number. – Rudy the Reindeer Sep 20 '12 at 15:56
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    Perhaps more relevant: http://math.stackexchange.com/questions/170334/whats-the-precise-meaning-of-imaginary-number/170342#170342 – Holdsworth88 Sep 20 '12 at 16:09
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    You might also like http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ – sakisk Sep 20 '12 at 19:42
  • ivan says that the set of imaginary numbers is $\mathbb{C}$, Matt (and wikipedia) says it is i$\mathbb{R}$. I've also heard $\mathbb{C} \setminus \mathbb{R}$. Whose right? – Michael Joyce Sep 20 '12 at 21:03
  • I don't think I ever understood them. I think it just tool to solve trig equations. – SamFisher83 Sep 20 '12 at 21:22
  • Quantum physics makes use of a variety of complex numbers, including a concept of imaginary *time*. Mind you, you'll still get real numbers when you do the mathematics that would result in most things that you'd expect a real number for (distance between two numbers on the complex plane is still a real number, as are the magnitudes of imaginary vectors). In a related manner, there are alternate formulations of quantum mechanics which involve negative probabilities instead, but I understand it makes the math more complex. Consider asking a physics-related stackexchange for more info. –  Sep 20 '12 at 21:52
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    Real numbers don't "exist" either, they're all just mathematicians' ideas. – akkkk Sep 20 '12 at 13:30
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    They can be used to represent/measure things that exist though. – NoChance Sep 20 '12 at 13:39
  • @EmmadKareem: What, like circles? – user1729 Sep 20 '12 at 13:44
  • One flavor of string theory predicts the Higgs Boson to be a particle with zero spin and imaginary mass (which is why we hadn't been able to find it, because we were looking for something with a real mass, because it gives objects mass therefore it must have some) – KeithS Sep 20 '12 at 14:50
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    a different way to put it is "real numbers are as imaginary as imaginary numbers". – mythealias Sep 20 '12 at 16:17
  • @user1729, In addition to the example provided by KeithS in his comment above, There are several examples, specially in the field of Electricity. See for example: http://www.math.toronto.edu/mathnet/questionCorner/complexinlife.html – NoChance Sep 20 '12 at 17:10
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    In what sense do ideas not exist? I would have said, for example, that while Santa Claus does not exist, the *idea* of Santa Claus certainly does exist. The widespread belief in God renders God an important *idea* that I should study and understand regardless of whether or not I believe that God actually exists; the *idea* of God certainly exists and has had a profound effect on the world. Or again: the only sense in which "love does not exist, it is only an idea in people's heads" is true is a silly and fatuous one. In short, I think your idea is wrong. – MJD Sep 20 '12 at 17:39
  • @MJD: this is exactly the confusion many people have with numbers, and which I am trying to point out, but it may be obvious that there is no way to measure love in the way you measure an electric potential. However, the /explanation/ of love is a very helpful one. – akkkk Sep 20 '12 at 21:21
  • @Auke If comlex numbers don't exist, please explain why the Schrödinger equation(the fundamental equation of quantum mechanics) uses them. I think that the use of complex numbers in the equation is essential, contrary to the use of them in, e.g. electronic circuits. – Makoto Kato Sep 21 '12 at 00:48
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    I think you just need to buy a better calculator. – icando Sep 21 '12 at 03:55
  • @EmmadKareem: Your comment was about $real numbers$ being used to measure things that exist, not about imaginary numbers. – user1729 Sep 21 '12 at 09:06
  • @MakotoKato: Schroedinger's equation does not exist in the physical sense, it's just an idea which apparently is very applicable to the measurable world. – akkkk Sep 21 '12 at 18:30
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    imaginary numbers have to be the single worst named concept in mathematics. I had some trouble with iuntil I got into upper level calc and engineering classes and realized my problem was all in the name. i is simply another dimension. – Mr.Mindor Sep 21 '12 at 18:47
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    "In mathematics, you don't understand things. You just get used to them. - John von Neumann – Paul Legato Sep 21 '12 at 19:55
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    @Auke Are you saying that every non-physical thing does not exit? – Makoto Kato Sep 21 '12 at 20:03
  • @MakotoKato: If "to exist" is defined in the sense that you can see or feel it, yes, but feel free to propose a clear definition of existence. However, my point is that imaginary numbers are just an idea, just like stories and limited liability companies, and in that sense they are just as "real" as integers. – akkkk Sep 21 '12 at 20:53
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    @Auke So an atom does not exit because we cannot see it nor can feel it. If it's just an idea like a story, how do you explain that the Schrödinger equation uses it? It describes the fundamental law of the universe. Without the complex numbers we cannot describe it. – Makoto Kato Sep 22 '12 at 01:37
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    @MakotoKato: I agree with you insofar that the idea of atoms is very real, and it very neatly describes the small thingies we're seeing through a microscope. I feel you are also completely missing my point, however. – akkkk Sep 22 '12 at 14:39
  • Relevant: http://math.stackexchange.com/questions/200776/can-i-keep-adding-more-dimensions-to-complex-numbers. – jcora Sep 22 '12 at 16:57
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    @akkkk "Real numbers don't "exist" either, they're all just mathematicians' ideas." This is exactly what I am disagreeing. The complex numbers were "discovered" in 16th century, long before they were used in the Schrödinger equation. Nobody dreamdt of quantum physics at that time. If they are just an idea like a story, how come they are used essentially in the equation which describes the fundamental law of the universe? – Makoto Kato Sep 22 '12 at 20:57
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    @MakotoKato: To answer this question, let me ask another question: All theories are written in some language, be it Latin, English, German or whatever. Using a language turns out to be essential for writing theories; you couldn't do it without. Should I conclude that languages are not a construct of the human mind, but somehow part of the intrinsic structure of the universe? – celtschk Sep 25 '12 at 06:48
  • @celtschk Mathematics is not a language. – Makoto Kato Sep 25 '12 at 07:27
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    @MakotoKato: I didn't say it is. I applied your *argument* to language, to demonstrate that it doesn't work. – celtschk Sep 25 '12 at 07:45
  • @celtschk Mathematics is not a language. So I don't see why your argument demonstrates that my argument does not work. – Makoto Kato Sep 25 '12 at 08:52
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    @MakotoKato: Your argument was using the rule: "X turns out to be very useful, if not essential, for theories of physics, therefore X is somehow part of the universe." You applied that argument for "X=mathematics". I applied it for "X=language" and got an obviously wrong result, thus invalidating the rule. Since the rule is invalid, its application to mathematics is also invalid. – celtschk Sep 25 '12 at 10:53
  • @celtschk It's not just useful. It's indispensable. If you don't know much about quantum physics, I guess it's hard for you to understand this, though. – Makoto Kato Sep 26 '12 at 21:13
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    @MakotoKato: I'm a physicist working in quantum information, so I know quantum mechanics *quite* well. However, I'm also able to distinguish between the model of reality (Quantum mechanics) and reality itself. The success of quantum mechanics shows that there's something in nature which can be quite well mapped onto mathematical structures using complex numbers. That does *not* imply that there are complex numbers in nature (just as the fact that for centuries we've successfully mapped real objects onto locations and momenta which are exactly defined at the same time doesn't imply that … – celtschk Sep 30 '12 at 11:10
  • there are in nature any objects which indeed do have well-defined locations and momenta; indeed, according to quantum mechanics, they don't, and modern experiments agree with the quantum mechanica predictions). – celtschk Sep 30 '12 at 11:12
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    @celtschk If the complex numbers(which were *discoverd* in 16th century) are just an idea like a story which one makes up freely, why is it used *essentially* to describe the fundamental law of the universe? Do you think it is an accident? – Makoto Kato Sep 30 '12 at 17:20
  • @celtschk Maybe the following link is of some use for you to understand what I'm talking about. http://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences – Makoto Kato Sep 30 '12 at 17:33
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    First: We *don't know* whether it is the fundamental law of the universe. It's the best we currently have. But then, classical mechanics was considered the fundamental laws for centuries, and today we know it isn't. Second: We use complex numbers for it because it turns out that they are *useful* in formulating the concepts which we use to model the universe. Note that we do *not* model the universe with complex numbers. We model the universe with operators and rays in Hilbert spaces. Now those Hilbert spaces are described with complex numbers. I don't know if the very same Hilbert spaces … – celtschk Sep 30 '12 at 17:37
  • could be described without complex numbers. However I know for sure that Hilbert spaces are not the only way to describe quantum mechanics (although I don't know enough about those alternative descriptions which have been developed to say whether they use complex numbers; but then, there are almost certainly others which we haven't yet found, and I see no reason to assume that all of them use complex numbers). – celtschk Sep 30 '12 at 17:40
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    @celtschk Even if it is not the fundamental law of the universe, the fact that the Schrödinger equation describes and predicts pyhsical phenomena *very* accurately remains. I repeat my question. Do you think it(that the complex numbers are used essentially in the equation) is an accident? – Makoto Kato Sep 30 '12 at 18:19
  • @MakotoKato: It might for example be an artefact of the fact that Lie groups are most easily worked with in their $GL(d,\mathbb C)$ representation. Also, if you look at the space of density matrices, they form a convex set in a *real* vector space (that's despite the density matrices are denoted by complex matrices). Unitary transformations translate into orthogonal transformations which keep that set invariant. And Hermitian operators are associated with linear function in that space. Who says that this real vector space is not more fundamental than the complex space of Hilbert space vectors? – celtschk Oct 04 '12 at 19:15
  • @MakotoKato: Just another argument which just came to me: You know that the Schrödinger equation has complex solutions. However if you look at the gauge transformations of the electromagnetic field and how they affect the wave function, you'll notice that you can completely gauge away the complex phases, and then get a real wave function and a corresponding electromagnetic potential and vector potential (if there were already such potentials to begin with, they are modified accordingly). That way, you get a completely real (but more complicated) form of quantum mechanics. – celtschk Oct 05 '12 at 08:01
  • @celtschk Are you trying to say that complex numbers are not indispensable in physics? – Makoto Kato Oct 05 '12 at 10:01
  • @MakotoKato: I can't tell for sure (for that, I'd have to know *much* more), but I definitely wouldn't exclude the possibility. – celtschk Oct 05 '12 at 10:40
  • @celtschk Do you agree with that complex numbers are indispensable in the Schrödinger equation? – Makoto Kato Oct 05 '12 at 22:36
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    @MakotoKato: That depends very much on what you mean with "the Schrödinger equation". If you mean the equation exactly as normally formulated, then yes, because there's an explicit i in it. But that doesn't tell us much because nature doesn't care about which of many equivalent ways to formulate a theory we use. – celtschk Oct 06 '12 at 21:06
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    @celtschk Another example: Complex analytic manifolds called Calabi–Yau manifolds play an important role in physics. Do you still think that it is an accident that complex numbers are used in physics? – Makoto Kato Oct 07 '12 at 17:32
  • Incidentally, "imaginary" numbers have physical existence (w/o even using quantum theory). See http://en.wikipedia.org/wiki/Phasor – Joshua Apr 15 '13 at 17:33
  • You only believe negative numbers "exist" because the idea has been presented to you since you were a small child and because you can see how naturally they fit in, i.e. even very simple math involving ones checking account might become "awkward" without them. But face it, you've never had negative one oranges in your hand or negative one of any physical object. Imaginary numbers are only imaginary in that they are so-named and in that they require more complicated math to recognize how terribly awkward it is to close one's eyes and pretend they don't exist. Insofar as -5 exists so does 5i. – John Robertson Mar 04 '14 at 06:42
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    @dustin Don't make gratuitious edits to old questions. – MJD Dec 19 '14 at 01:16
  • @MJD it wasn't gratuitous. The words the square root of -1 one has a well understood symbol and notation which is $i=\sqrt{-1}$. – dustin Dec 19 '14 at 01:19
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    @dustin Changing it does not add any clarity or correctness to the question, which was unnecessarily bumped to the front page. – MJD Dec 19 '14 at 04:30
  • When it is not possible to take sqrt , then to proceed even with that uncertainty, mathematicians took concept of imagery. – Anurag Nov 06 '16 at 08:01
  • I think this question has a really good answer at https://math.stackexchange.com/questions/154/do-complex-numbers-really-exist – Timothy Dec 17 '18 at 03:55

22 Answers22


Let's go through some questions in order and see where it takes us. [Or skip to the bit about complex numbers below if you can't be bothered.]

What are natural numbers?

It took quite some evolution, but humans are blessed by their ability to notice that there is a similarity between the situations of having three apples in your hand and having three eggs in your hand. Or, indeed, three twigs or three babies or three spots. Or even three knocks at the door. And we generalise all of these situations by calling it 'three'; same goes for the other natural numbers. This is not the construction we usually take in maths, but it's how we learn what numbers are.

Natural numbers are what allow us to count a finite collection of things. We call this set of numbers $\mathbb{N}$.

What are integers?

Once we've learnt how to measure quantity, it doesn't take us long before we need to measure change, or relative quantity. If I'm holding three apples and you take away two, I now have 'two fewer' apples than I had before; but if you gave me two apples I'd have 'two more'. We want to measure these changes on the same scale (rather than the separate scales of 'more' and 'less'), and we do this by introducing negative natural numbers: the net increase in apples is $-2$.

We get the integers from the naturals by allowing ourselves to take numbers away: $\mathbb{Z}$ is the closure of $\mathbb{N}$ under the operation $-$.

What are rational numbers?

My friend and I are pretty hungry at this point but since you came along and stole two of my apples I only have one left. Out of mutual respect we decide we should each have the same quantity of apple, and so we cut it down the middle. We call the quantity of apple we each get 'a half', or $\frac{1}{2}$. The net change in apple after I give my friend his half is $-\frac{1}{2}$.

We get the rationals from the integers by allowing ourselves to divide integers by positive integers [or, equivalently, by nonzero integers]: $\mathbb{Q}$ is (sort of) the closure of $\mathbb{Z}$ under the operation $\div$.

What are real numbers?

I find some more apples and put them in a pie, which I cook in a circular dish. One of my friends decides to get smart, and asks for a slice of the pie whose curved edge has the same length as its straight edges (i.e. arc length of the circular segment is equal to its radius). I decide to honour his request, and using our newfangled rational numbers I try to work out how many such slices I could cut. But I can't quite get there: it's somewhere between $6$ and $7$; somewhere between $\frac{43}{7}$ and $\frac{44}{7}$; somewhere between $\frac{709}{113}$ and $\frac{710}{113}$; and so on, but no matter how accurate I try and make the fractions, I never quite get there. So I decide to call this number $2\pi$ (or $\tau$?) and move on with my life.

The reals turn the rationals into a continuum, filling the holes which can be approximated to arbitrary degrees of accuracy but never actually reached: $\mathbb{R}$ is the completion of $\mathbb{Q}$.

What are complex numbers? [Finally!]

Our real numbers prove to be quite useful. If I want to make a pie which is twice as big as my last one but still circular then I'll use a dish whose radius is $\sqrt{2}$ times bigger. If I decide this isn't enough and I want to make it thrice as big again then I'll use a dish whose radius is $\sqrt{3}$ times as big as the last. But it turns out that to get this dish I could have made the original one thrice as big and then that one twice as big; the order in which I increase the size of the dish has no effect on what I end up with. And I could have done it in one go, making it six times as big by using a dish whose radius is $\sqrt{6}$ times as big. This leads to my discovery of the fact that multiplication corresponds to scaling $-$ they obey the same rules. (Multiplication by negative numbers responds to scaling and then flipping.)

But I can also spin a pie around. Rotating it by one angle and then another has the same effect as rotating it by the second angle and then the first $-$ the order in which I carry out the rotations has no effect on what I end up with, just like with scaling. Does this mean we can model rotation with some kind of multiplication, where multiplication of these new numbers corresponds to addition of the angles? If I could, then I'd be able to rotate a point on the pie by performing a sequence of multiplications. I notice that if I rotate my pie by $90^{\circ}$ four times then it ends up how it was, so I'll declare this $90^{\circ}$ rotation to be multiplication by '$i$' and see what happens. We've seen that $i^4=1$, and with our funky real numbers we know that $i^4=(i^2)^2$ and so $i^2 = \pm 1$. But $i^2 \ne 1$ since rotating twice doesn't leave the pie how it was $-$ it's facing the wrong way; so in fact $i^2=-1$. This then also obeys the rules for multiplication by negative real numbers.

Upon further experimentation with spinning pies around we discover that defining $i$ in this way leads to numbers (formed by adding and multiplying real numbers with this new '$i$' beast) which, under multiplication, do indeed correspond to combined scalings and rotations in a 'number plane', which contains our previously held 'number line'. What's more, they can be multiplied, divided and rooted as we please. It then has the fun consequence that any polynomial with coefficients of this kind has as many roots as its degree; what fun!

The complex numbers allow us to consider scalings and rotations as two instances of the same thing; and by ensuring that negative reals have square roots, we get something where every (non-constant) polynomial equation can be solved: $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$.

[Final edit ever: It occurs to me that I never mentioned anything to do with anything 'imaginary', since I presumed that Sachin really wanted to know about the complex numbers as a whole. But for the sake of completeness: the imaginary numbers are precisely the real multiples of $i$ $-$ you scale the pie and rotate it by $90^{\circ}$ in either direction. They are the rotations/scalings which, when performed twice, leave the pie facing backwards; that is, they are the numbers which square to give negative real numbers.]

What next?

I've been asked in the comments to mention quaternions and octonions. These go (even further) beyond what the question is asking, so I won't dwell on them, but the idea is: my friends and I are actually aliens from a multi-dimensional world and simply aren't satisfied with a measly $2$-dimensional number system. By extending the principles from our so-called complex numbers we get systems which include copies of $\mathbb{C}$ and act in many ways like numbers, but now (unless we restrict ourselves to one of the copies of $\mathbb{C}$) the order in which we carry out our weird multi-dimensional symmetries does matter. But, with them, we can do lots of science.

I have also completely omitted any mention of ordinal numbers, because they fork off in a different direction straight after the naturals. We get some very exciting stuff out of these, but we don't find $\mathbb{C}$ because it doesn't have any natural order relation on it.

Historical note

The above succession of stages is not a historical account of how numbers of different types are discovered. I don't claim to know an awful lot about the history of mathematics, but I know enough to know that the concept of a number evolved in different ways in different cultures, likely due to practical implications. In particular, it is very unlikely that complex numbers were devised geometrically as rotations-and-scalings $-$ the needs of the time were algebraic and people were throwing away (perfectly valid) equations because they didn't think $\sqrt{-1}$ could exist. Their geometric properties were discovered soon after.

However, this is roughly the sequence in which these number sets are (usually) constructed in ZF set theory and we have a nice sequence of inclusions $$1 \hookrightarrow \mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$

Stuff to read

  • The other answers to this question give very insightful ways of getting $\mathbb{C}$ from $\mathbb{R}$ in different ways, and discussing how and why complex numbers are useful $-$ there's only so much use to spinning pies around.
  • A Visual, Intuitive Guide to Imaginary Numbers $-$ thanks go to Joe, in the comments, for pointing this out to me.
  • Some older questions, e.g. here and here, have some brilliant answers.

I'd be glad to know of more such resources; feel free to post any in the comments.

Clive Newstead
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    Excellent effort. I just want to know what is the relationship between multiplication and rotation? I mean why is 1 rotation = i and 2 rotations = i*i? – NoChance Sep 20 '12 at 13:35
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    I registered to math.stackexchange just to vote for this wonderful answer! – lukas.pukenis Sep 20 '12 at 13:35
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    @Emmad: If you have a point $z$ on the complex plane, then $iz$ is what you get by rotating it by $90^{\circ}$ clockwise about the origin. And $-z=i^2z=i \cdot iz$ is what you get by rotating $z$ by $180^{\circ}$ about the origin. – Clive Newstead Sep 20 '12 at 13:39
  • @lukas.pukenis: Ačiū! – Clive Newstead Sep 20 '12 at 13:39
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    This is probably the best plain-English, grade-school-level explanation of the various sets of numbers I have ever heard. WAY better than anything my teachers on the subject could come up with. – KeithS Sep 20 '12 at 14:42
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    +1 but you didn't complete the definition that the `imaginary number` line as simply the axis orthogonal to the `real number` line in the `complex number` plane so that every `complex number` can be expressed as the sum of a `real number` and an `imaginary number`. – StarNamer Sep 20 '12 at 14:59
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    Excellent explanation! The first time I've seen the analogy of rotation was here, which also contains a really good, visual exploration of imaginary numbers: http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/ – Joe Sep 20 '12 at 15:01
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    @StarNamer: No, but did I need to? I know the question asks about 'imaginary numbers', but I think what he meant was complex numbers on the whole. – Clive Newstead Sep 20 '12 at 15:01
  • @Joe: That's very cool! – Clive Newstead Sep 20 '12 at 15:08
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    @CliveN. Man! This is one of the greatest, clearest, coolest and simplest answers I've ever seen for this question (and funniest, I had a lot of fun reading it). I will quote it every time I get the chance! +1 (If I could vote more than one time, I would) – Barranka Sep 20 '12 at 15:52
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    +1 This has to be the erudite, but relevant explanation of this I've seen yet :D – Sinaesthetic Sep 20 '12 at 16:09
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    Very clear, and incidentally shows most clearly that multiplication is, after all, repeated addition. – boisvert Sep 20 '12 at 23:15
  • @CliveN Possiby not. Your answer is probably the clearest brief history of the evolution of number systems I've seen. However you did mention that the `'number plane' ... contains our previously held 'number line'` and that would have been a good place to mention it. – StarNamer Sep 21 '12 at 00:54
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    Oh, please, please, please, expand your answer to quaternions, please! – Dan Ganiev Sep 21 '12 at 07:12
  • @Joe, and OP: That article could probably go in the answer. It is a great resource. – George Duckett Sep 21 '12 at 07:20
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    I'm sorry to spoil the consensus, but while the first 4 sections were great (apart from the unexplained "$\mathbb{Z}$ is the closure of $\mathbb{N}$ under the operation $-$" etc), the actual imaginary numbers part of this left me with more questions than answers. Why a quarter turn? Why $i^4=1$ and not $4i=1$? What does it mean for -1 to be a half-turn, if that's why we're rejecting 1? What does this have to do with any other use of imaginary numbers? Where are the complex numbers? – IMSoP Sep 21 '12 at 07:32
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    @DanielExcinsky: Also octonions... – Richie Cotton Sep 21 '12 at 09:38
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    Wow, I can't say I was expecting this answer to get anywhere near this many up-votes, I kind of just scribbled it down... I'll make some edits to respond to the above comments, thanks for the feedback! – Clive Newstead Sep 21 '12 at 11:06
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    With the integers, debt was born. I wish my bank to forget the integers when billing me. – Red Banana Sep 21 '12 at 12:06
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    OMG! [Hamilton](http://en.wikipedia.org/wiki/William_Rowan_Hamilton) was an alien! I knew it! I knew it all along! – Red Banana Sep 21 '12 at 12:26
  • BTW, the sequence is wrong, this is the right one: $$Steve\;Jobs \hookrightarrow Apples \hookrightarrow 1 \hookrightarrow \mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow Pies \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$ – Red Banana Sep 21 '12 at 12:32
  • I would only add that complex numbers are rotations because of a nice fact about complex numbers: if you do a scaled rotation of a 2D point $[x, y]$ you get the ugly point $[x r \cos(t) - y r \sin(t), y r \cos(t) + x r \sin(t)]$. This looks cleaner if we define $a = r \cos(t)$, $b = r \sin(t)$, as $[a x - by, a y + b x]$. Perhaps surprisingly, we can write this same outcome as $(a + i b) (x + i y) = (a x - b y) + i (a y + b x)$. It's not so surprising: normal multiplication becomes addition via logarithms; and rotations are matrix multiplications which become additions: $R(s) R(t) = R(s + t)$. – CR Drost Sep 21 '12 at 18:58
  • @ChrisDrost: Indeed; for my purposes, the polar form of the complex number is a much cleaner way of doing things. I didn't want to go into too much detail with it, which is why I glossed over so much. But thanks for your comment! – Clive Newstead Sep 21 '12 at 19:05
  • Explaining complex numbers via scaling and rotation - transformations of the plane - is a great approach. This opens an opportunity to mention that a lot of advanced mathematics is about transformation rules for more complicated kinds of objects and data. (That the usage of the term `number' ends with quaternions and octonions is a rather arbitrary historical accident.) – Bob Pego Sep 22 '12 at 02:28
  • I wish roots, e.g. $\sqrt{2}$ would have been left out of this. I think you did it because you needed it to express radius (instead of ratio) - which was unnecessary relative to scale or rotations. It did let you imply $\sqrt{-1}$ but your polynomial teaser was more satisfying. – New Alexandria Sep 22 '12 at 12:56
  • @EmmadKareem Each of the "rotation" steps is obvious when examined separately: i*i = -1 (by definition), i*-1 = -i (obviously), and i*-i = 1 (since i*i = 1) - this also true when applied to any other particular complex number. – Random832 Sep 24 '12 at 16:06
  • s/measily/measly – g33kz0r Sep 24 '12 at 21:07
  • @Random832, thanks for the explanation. – NoChance Sep 24 '12 at 22:17
  • @StarNamer: I have at least given in and mentioned imaginary numbers. – Clive Newstead Sep 25 '12 at 11:35
  • You seem to have been persuaded to go well beyond the question and into the realms of hypercomplex numbers. While everyone else seems to have enjoyed the comments & extensions, I haven't seen if @SachinKainth found the answer helpful! – StarNamer Sep 25 '12 at 16:56
  • @StarNamer: I added all the extra stuff (incl. quaternions, octonions, etc) upon request *after* Sachin had accepted the answer, so I hope it didn't matter too much. I hope I haven't gone as far as to confuse anyone by adding the extra stuff in (I was already quite reluctant to mention quaternions etc). – Clive Newstead Sep 25 '12 at 17:02
  • Just to add, the 'i' is of a great physical significance in Electrical circuits eg- it helps to find the lag/lead of Currents or Voltages across certain elements. – Ishank Sep 26 '12 at 07:24
  • "[. . .]divide integers by positive integers [or, equivalently, by nonzero integers][. . .]" Nonzero and positive are not equivalent adjectives for integers. Nonzero integers means $\{\pm 1, \pm 2, \pm 3,\dots\}$ whereas positive integers means $\{1,2,3,\dots\}$. – 000 Sep 26 '12 at 21:26
  • @Limitless: That's not what I meant. We can define the rationals as either 'integers divided by positive integers' or 'integers divided by nonzero integers' and the two definitions are equivalent: if $\frac{p}{q}$ is a rational with $p \in \mathbb{Z}$ and $q<0$ then $\frac{-p}{-q}$ is rational (and $-p \in \mathbb{Z}$ and $-q > 0$). That is, every rational can be written as an integer divided by a positive integer. – Clive Newstead Sep 26 '12 at 23:50
  • What an excellent answer ;). Thanks a bunch. – Chan Sep 30 '12 at 04:12
  • @CliveN., oh! Thank you for that clarification. – 000 Sep 30 '12 at 15:02
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    I only read through the first two when I upvoted this. I could see where it was going from there. – colboynik Feb 20 '13 at 19:56
  • I logged in not only to upvote this, but also to favorite it because this answer is so awesome that it makes me want to learn more. I used to be a somewhat good math student in middle school, but we sure didn't learn this stuff - and as a consequence, I've lost a lot of my knowledge. (I remember dealing with i, but I never knew what it was _for_ - I didn't realize that it could be applied to rotations around the unit circle!) Seeing, again, how the sets of numbers follow from each other makes me want to pick this all up again. Thank you! – lunchmeat317 Mar 20 '14 at 00:08
  • A small note... When it comes to counting numbers we get $\mathbb{N}^+$, as we cannot count to $0$. Counting only yields $1,2,3,4,\cdots$ The number $0$ is already a concept based on counting... – johannesvalks Jun 22 '14 at 21:17
  • So evolution or blessing? – Peter Jan 20 '15 at 03:27
  • "humans are blessed by their ability to notice that there is a similarity between the situations of having three apples in your hand and having three eggs in your hand". So simple, so powerful. If only people sat and thought about such fundamental things... – Frank Bryce Aug 08 '15 at 03:30
  • And if I flip the pie upside down? – Simply Beautiful Art Apr 23 '16 at 00:58
  • So... why are the numbers **imaginary**? They should be called, instead, the **lateral numbers**, regardless if "they don't exist" $-$ they do, but not on the one-dimensional number-line we are all familiar with. (Yes, that's right: numbers are *not* one dimensional.) – Mr Pie May 11 '18 at 15:25
  • kudos to you. explained with very good examples. i loved the way you progressively explained from easy to hard concepts. – Parag Meshram Aug 23 '19 at 16:29
  • We gave up at the "real numbers" part because we had to get on with our lives.That's where our things got sour with mathematics. – user117 Jun 19 '20 at 10:39
  • but I don't understand the geometric meaning of $\color\red{adding}$ $i$ can you help me explain? – Soumyadip Mukherjee Jan 01 '21 at 06:53
  • @inquisitivesoumyadip: Addition of complex numbers corresponds with translation in the complex plane. The pie analogy isn't as useful for understanding this, though. – Clive Newstead Jan 02 '21 at 18:04

You ask why imaginary numbers are useful. As with most extensions of number systems, historically such generalizations were invented because they help to simplify certain phenomena in existing number systems. For example, negative numbers and fractions permit one to state in a single general form the quadratic equation and its solution (older solutions bifurcated into many cases, avoiding negative numbers and fractions). One of the primary reasons motivating the invention of complex numbers is that they serve to linearize what would otherwise be nonlinear phenomena - thus greatly simplifying many problems. Here are some examples.

Consider the problem of representing integers as sums of squares $\rm\: n = x^2 + y^2$. Early solutions to this and related problems employed a complicated arithmetic of binary quadratic forms. Such arithmetic was quite intricate and often very nonintuitive, e.g. even the proof of associativity of composition of such forms was a tour de brute force, occupying pages of unmotivated computations in Gauss' Disq. Arith. But this quadratic arithmetic of binary quadratic forms can be linearized. Indeed, by factorization $\rm\: x^2 + y^2 = (x+y{\it i})(x-y{\it i}),$ which allows us to view sums of squares as norms of Gaussian integers $\rm\:x+y{\it i},\ \ x,y\in \Bbb Z.\:$ But just like the rational integers $\Bbb Z,$ these "imaginary" integers have a Euclidean algorithm, so enjoy unique factorization into primes. By considering all the possible factorizations of $\rm\:n\:$ in the Gaussian integers we obtain all the possible representations of $\rm\:n\:$ as a sum of squares. In a similar way, "rational, real" arithmetic of integral quadratic forms becomes much simpler by passing to the "irrational" and/or "imaginary" arithmetic of quadratic number fields. This line of research led to the discovery of ideals and modules, fundamental linear structures at the heart of modern number theory and algebra. [See this answer and its links for a more precise description of the equivalence between quadratic forma and ideals].

Thus, by factorizing completely over $\Bbb C$, we have reduced the complicated nonlinear arithmetic of binary quadratic forms to the simpler, linear arithmetic of Gaussian integers, i.e. to the more familiar arithmetical structure of a unique factorization domain (in fact a Euclidean domain). Analogous linearization serves to simplify many problems. For example, when integrating or summing rational functions (quotients of polynomials), by factoring denominators over $\Bbb C$ (vs. $\Bbb R)$ and taking partial fraction decompositions, the denominators are at worst powers of linear (vs. quadratic) polynomials - which greatly simplifies matters. More generally, when solving constant coefficient differential or difference equations (recurrences), by factoring their characteristic (operator) polynomials over $\Bbb C,$ we reduce to solutions of linear (vs. quadratic) differential or difference equations. In the same way, there are many real problems (over $\Bbb R)$ whose simplest solutions are obtained by an imaginary detour (over $\Bbb C).$ Perhaps readers will mention more such problems in the comments.

Bill Dubuque
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I went to school for electrical engineering ($7$ years total) and we used imaginary numbers all over the place.

Even with all that schooling, this is probably the clearest explanation of imaginary numbers I've seen: A Visual, Intuitive Guide to Imaginary Numbers.


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    I especially loved the historical note showing how offended people in the mid-1700's were by negative numbers... it clarifies the fact that "imaginary" and "negative" are just labels for parts of the plane. – Jerry Andrews Sep 20 '12 at 21:22
  • I had seen this article before, and I must agree -- IT IS EXCELLENT. If you're grappling with the concept of imaginary numbers, you MUST check it out! – Charlie Flowers Sep 21 '12 at 18:15
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    That article is great! This should've been the accepted answer! – B Faley Sep 24 '12 at 11:13
  • This link was my first thought - completely worth the time to read – Deebster Sep 24 '12 at 11:55

Well, as you know there's no real number whose square is negative. But now imagine numbers which are. Let's call them imaginary. Now what properties would such numbers have? Well, there would be for example a number whose square is $-1$. Let's call that number the imaginary unit and give it the name $\mathrm i$. Now if we multiply this number with some real number, that is, use $r\mathrm i$, we get a number whose square is $(\mathrm ir)^2 = \mathrm i^2r^2 = -r^2$. Since all positive numbers can be written as $r^2$, we get that all negative numbers can be written as $(\mathrm ir)^2$. Thus the products $\mathrm ir$ are our imaginary numbers. We also see that $(-\mathrm i)^2 = (-1)^2\mathrm i^2 = -1$, so there are actually two numbers whose square is $-1$ (which makes sense because, after all, there are also two numbers whose square is $1$, namely $1$ and $-1$).

OK, but what happens if we add a real number and one of out imaginary numbers. Well, now things get complex. We get general complex numbers.

OK, but how do we know that we've not just made some nonsense, similar to the nonsense that we get when we invent a number $o$ so that $0o=1$? Well to see that, we recognize that all complex numbers are of the form $x+\mathrm iy$ with real numbers $x$ and $y$, and thus the pair $(x,y)$ completely specifies a complex number. Therefore now we re-derive the complex numbers as pairs of real numbers, but now using proper mathematical instruments so we know for sure that whatever we do is well defined. Since doing that we arrive at the very same structure which we just had derived in a quite informal way, we know that the complex numbers are a sound mathematical structure.

OK, now that we have invented the imaginary and complex numbers, are they useful for something? Well, indeed they are. For example, several mathematical statements are much easier in complex numbers than in real numbers. For example, with complex numbers, every polynomial can be written in the form $a(x-x_1)(x-x_2)\cdots(x-x_n)$. With real numbers, this is impossible for polynomials having for example factors of the form $(x^2+1)$. Moreover, we have the very useful relation $\mathrm e^{\mathrm i\phi} = \cos\phi + \mathrm i\sin\phi$. So forget about complicated addition theorems for sine and cosine. Just rewrite your formula in complex exponentials and enjoy the simple relation $\mathrm e^{\mathrm i(\alpha+\beta)}=\mathrm e^{\mathrm i\alpha}\mathrm e^{\mathrm i\beta}$.

Finally, if you want to do quantum physics (and almost all modern physics is quantum physics) you'll find that you have to use complex numbers.

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The term "imaginary" is somewhat disingenuous. It's a real concept, with real (at least theoretical) application, just like all the "real" numbers.

Think back to that algebra class. You were asked to solve a polynomial equation; that is, find all the values of X for which the entire equation evaluates to zero. You learned to do this by polynomial factoring, simplifying the equation into a series of first-power terms, and then it was easy to see that if any one of those terms evaluated to zero, then everything else, no matter its value, was multiplied by zero, producing zero.

You tried this on a few quadratic equations. Sometimes you got one answer (because the equation was $y=ax^2$ and so the only possible answer was zero), sometimes you got two (when the equation boiled down to $y= (x\pm n)(x \pm m)$, and so when $x=-m$ or $x=-n$ the equation was zero), and a couple of times, you got no answers at all (usually, an equation that breaks down to $y=(x+n)(x+m)$ doesn't evaluate to zero at $x=-m$ or $x=-n$).

In your algebra class, you're told this just happens sometimes, and the only way to make sure any factored term $(x\pm k)$ represents a real root is to plug in $-k$ for $x$ and solve. But, this is math. Mathematicians like things to be perfect, and don't like these "rules of thumb", where a method works sometimes but it's really just a "hint" of where to look. So, mathematicians looked for another solution.

This leads us to application of the quadratic formula: for $ax^2 + bx + c = 0$, $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. This formula is quite literally the solution of the general form of the equation for x, and can be derived algebraically. We can now plug in the coefficients, and find the values of $x$ where $ax^2 + bx + c=0$. Notice the square root; we're first taught, simply, that if $b^2-4ac$ is ever negative, then the roots you'd get by factoring the equation won't work, and thus the equation has no real roots. $b^2-4ac$ is called the determinant for this reason.

But, the fact that $b^2-4ac$ can be negative remains a thorn in our side; we want to solve this equation. It's sitting right in front of us. If the determinant were positive, we would have solved it already. It's that pesky negative that's the problem.

Well, what if there was something we could do, that conforms to the rules of basic algebra, to get rid of the negative? Well, $-m = m*-1$, so what if we took our term that, for the sake of argument, evaluated to $-36$, and made it $36*-1$? Now, because $\sqrt{mn} = \sqrt{m}\sqrt{n}$, $\sqrt{-36} = \sqrt{36}\sqrt{-1} = 6\sqrt{-1}$. We've simplified the expression by removing what we can't express as a real number from what we can.

Now to clean up that last little bit. $\sqrt{-1}$ is a common term whenever the determinant is negative, so let's abstract it behind a constant, like we do $\pi$ and $e$, to make things a little cleaner. $\sqrt{-1} = i$. Now, we can define some properties of $i$, particularly a curious thing that happens as you raise its power:

$$i^2 = \sqrt{-1}^2 = -1$$ $$i^3 = i^2*i = -i$$ $$i^4 = i^2*i^2 = -1*-1 = 1$$ $$i^5 = i^4*i = i$$

We see that $i^n$ transitions through four values infinitely as its power $n$ increases, and also that this transition crosses into and then out of the real numbers. Seems almost... circadian, rotational. As Clive N's answer so elegantly explains it, that's what imaginary numbers represent; a "rotation" of the graph through another plane, where the graph DOES cross the $x$-axis. Now, it's not actually really a circular rotation onto a new linear z-plane. Complex numbers have a real part, as you'd see by solving the quadratic equation for a polynomial with imaginary roots. We typically visualize these values in their own 2-dimensional plane, the complex plane. A quadratic equation with imaginary roots can thus be thought of as a graph in four dimensions; three real, one imaginary.

Now, we call $i$ and any product of a real number and $i$ "imaginary", because what $i$ represents doesn't have an analog in our "everyday world". You can't hold $i$ objects in your hand. You can't measure anything and get $i$ inches or centimeters or Smoots as your result. You can't plug any number of natural numbers together, stick a decimal point in somewhere and end up with $i$. $i$ simply is.

As far as having use outside "ivory tower" math disciplines, a big one is in economics; many economies of scale can be described as a function of functions of the number of units produced, with a cost term and a revenue term (the difference being profit or loss), each of these in turn defined by a function of the per-unit sale price or cost and the number produced. This all generally simplifies to a quadratic equation, solvable by the quadratic formula. If the roots are imaginary, so are the breakeven points (and your expected profits).

Another good one is in visualizations of complex numbers, and of their interactions when multiplied. The first one I was exposed to is a well-known series set, produced by taking an arbitrary complex number, squaring it ($(a+bi)^2 = (a+bi)(a+bi) = a^2 + 2abi + b^2i^2 = a^2-b^2 + 2abi$), and then adding back its original value. Repeated to infinity with this number, the series either converges to zero or diverges to infinity (with a few starting numbers exhibiting periodicity; they'll jump around infinitely between a finite number of points much like $i$ itself does). The set of all complex numbers for which the series does not diverge is the Mandelbrot set or M-set, and while the area of the graph is finite, its perimeter is infinite, making the graph of this set a fractal (one of the most highly-studied, in fact).

The Mandelbrot set can in turn be defined as the set of all complex numbers $c$ for which the Julia set $J(f)$ of $f(z)=z^2 + c \to z$ is connected. A Julia set exists for every complex polynomial function, but usually the most interesting and useful sets are the ones for values of $c$ that belong to the M-set; Julia fractals are produced much the same way as the M-set (by repeated iteration of the function to determine if a starting $z$ converges or diverges), but $c$ is constant for all points of the set instead of being the original point being tested. You can define Julia sets with all sorts of fractal shapes. These fractals, more accurately the iterative evaluation behind them, are used for pseudorandom number generation, computer graphics (the sets can be plotted in 3-d to create landscapes, or they can be used in shaders to define complex reflective properties of things like insect shells/wings), etc.

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This question has already been answered quite thoroughly, but I just want to add that generally speaking, besides the whole numbers, none of the numbers we use "exist" in the real world. The only reason we have adopted extensions to the whole numbers to the natural, integer, rational, real, and complex sets in turn is because these extensions make problems solvable when thinking abstractly. At the end of the day, everything relates back to the whole numbers, however.

Most people use all of the sets except for complex numbers in very commonplace, everyday situations, which is why we've come to view everything up to the real numbers as being fairly intuitive, at least at first glance. (When you dig under the surface, everything gets a great deal more subtle, which is why there are people who study primarily numbers, who we call number theorists. But that's a whole other story.)

It's important to note that this progression isn't the only way to extend the whole numbers. There are hundreds of different arithmetics that have been designed, many not even based on the whole numbers. It's just that the usual extension applies to so many situations that come up commonly. (People who study universal algebra study the ways in which different possible math systems are alike and different. But that's a whole other story as well.)

Complex numbers have taken their place as the normal extension to the reals because they are so useful when dealing with polynomials, which happen to arise in a massive number of mathematical situations. They also allow the exponential function and the trigonometric functions to be viewed as special cases of the same thing, through Euler's Formula, which enables all sorts of great algebra tricks. Specifically, these sorts of functions pop up constantly when using either Taylor or Fourier series to simplify the process of working on problems with tricky transcendental functions. Complex numbers make dealing with these representations a breeze (relatively).

There are even further extensions. If instead of worrying about how to take the square root of -1, you worry about what happens past infinity, the real numbers can alternatively be expanded in several jumps to include hyperreals, superreals, and surreals. None of these systems have caught on, though, because we have alternative ways of dealing with the infinite and infinitesimal quantities in calculus that people find more powerful/convenient.

You can also zip on past complex numbers to Quaternions, and octonions on top of that. Vectors generalize all of the above. They aren't often though of as numbers, but are similar in that they generalize the concept of a property of an object having a mathematical value. Matrixes generalize vectors, and tensors generalize matrixes.

As you climb this ladder, you gain more and more mathematical power, but you start to lose properties that we expect of whole numbers. For complex numbers, order (greater than/less than) begins to become ambiguous. We generally don't think of vectors as "numbers" because we want all operations on vectors to work regardless of dimension, and most of the arithmetic operations don't really generalize. With matrixes, the commutative property goes out the window, and things start to get really weird, especially when the matrixes aren't square. And so forth.

All of this to say that numbers are best viewed as machinery. Different number systems are really only used to the extent which they make a given math situation or problem easier to think about. If you're an engineer, complex numbers do this in many, many situations, which justifies their added...complexity. If you're not an engineer, they're definitely worth understanding, but you may not find uses for them on a daily basis.

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  • Is there a page comparing the number, complex number, quaternion, octonion, vector, matrix, and tensor solutions of a problem or two? Are tensors the last thing one would need to know or is there something even better? Like the myriad of JavaScript frameworks, i'd like the simplest and most powerful general option. – Cees Timmerman May 27 '18 at 15:25
  • @CeesTimmerman There is no endpoint. There are only ever more abstract abstractions. Pure mathematicians study these things for their own sakes, looking for deeper truths, while applied mathematicians develop the math around use cases. Much like a programming framework, you reach for something that has the right combination of overhead to learn and use versus the benefit you get. You might choose a more familiar tool, even if its not the optimal fit. This is one reason why matrices are so popular. – acjay Jun 01 '18 at 02:00

Complex numbers are just a handy way to handle two dimensional points and move them around. The key to it is understanding that i × i = −1 is just a simple by-product of moving these points around.

Real numbers correspond to numbers on a line (one dimension), which is usually how they are represented: a single axis where each number has a position. Operations on these real numbers have been defined to apply the two most basic transformations:

  • Translation (addition)

    Move a point by a given amount.

  • Scaling (multiplication)

    Move a point by an amount related to its value, e.g., two times further than it was.

Now, for a number of situations, you need to handle elements that are not on a line, but on a plane—you are now in two dimensions. When working in two dimensions, you need to know where you are horizontally and vertically, which you usually represent with two numbers. For instance, (3, 2) is “3 to the right, 2 up”. Complex numbers are designed to manipulate these elements with dimensions with “simple” mathematics.

We define i as being the vertical unit. 2i is “2 up”, −4i is “4 down”, and 3 + 2i is “3 to the right and 2 up”. We still can use translation and scaling like in the one-dimension case, but we would like to add something: rotation. How do I turn “2 to the right” into “2 up”?

The solution comes with multiplying by i. If 1 is “1 to the right” and 1 × i is “1 up”, then it means that multiplying by i is simply rotating by 90 degrees with point 0 as a center, counter-clockwise. 2 × i = 2i means “2 to the right” multiplied by i gives “2 up”.

And this is where it gets interesting: rotating the point “1 to the right” by 90 degrees gives “1 up”. Rotating it again by 90 degrees gives “1 left”. This means that multiplying 1 twice by i gives −1.

We have 1 × i × i = −1, and since i × i = −1, i is by definition the square root of −1.

Jon Purdy
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    "Complex numbers are just a handy way to handle two dimensional points and move them around." No, it's not that simple. The Schrödinger equation(the fundamental equation of quantum mechanics) uses them. Its use is essential, contrary to the use of them in, e.g. electronic circuits. IMO, complex numbers are indispensable to describe the laws of physics. – Makoto Kato Sep 21 '12 at 05:12

This argument is a loose argument for the sake of simplicity and because I know little about the subject. However, I think it may be good for non-mathematicians.

The simplistic view is to note that imaginary numbers (or Complex Numbers) are numbers that are defined by humans to describe quantities different from the numbers we use in our day-to-day life (unless you are a scientist). They have certain rules that are somewhat different than those we use to calculate with non-complex numbers. Hence, the subjects of (Complex Variables and Complex Analysis).

In mathematics, this is not strange. There are concepts that may look surprising until you study them carefully. For example, in Binary Numbers $1+1=10$. This result does not make any sense unless you understand and realize that the result is valid in the Binary System, domain or framework.

Personally, I thought about this before I read your question, and found that the problem comprehending such concepts could arise when you think about a concept outside its framework (or domain) and try to rationalize the results using our every day concepts.

For example trying to evaluate the $\sqrt{-1}$ on a regular calculator with no setting for Imaginary Arithmetic (the proper name is probably Complex Arithmetic). The calculator has to be set to the correct mode (or framework) to give a correct result. In fact, the software in your calculator should have given you a decent error message (or better yet the result of $i$ with a warning note).

Again, the same thing will happen if you are using your calculator in Binary mode to add $1+1$, you will not get the familiar $2$.

Many other examples can be driven around the same concept.

I hope this helps.

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I think of i as just a symbol to represent an operation


When we want the square root of -1, just represent the whole statement with a symbol without evaluating it. This avoids the necessity of trying to explain it further, we don't need to map the answer to some real world concept, it's just a saved operation. We also know that the square root has the following property:

      √x √x = x

No matter what the x. i.e.

      √-1 √-1 = -1

Numbers are useful to me when they represent concepts in the real world. I don't map i to anything in the real world but with this ability to represent the operation, I can now manipulate it in algebraic expressions to ultimately get back to non-imaginary numbers that I do find useful. http://en.wikipedia.org/wiki/Euler%27s_formula

Riaz Rizvi
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Check it out, I just learned this very recently:

Define the set of all ordered pairs $(x, y)$, call it $\mathbb{C}$, the set of complex numbers. We call $x$ the real part, and y the imaginary part. Now define multiplication like this:

$(x, y) \cdot (a,b) = (xa-yb, xb+ya) $

Now I'm not sure what that's supposed to be but observe:

$(0, 1)^2 = (0,1) \cdot(0,1)= (0-1, 0+0) = (-1, 0)$

Since the second number in the ordered pair is the imaginary part, (0,1) corresponds to $0+1\cdot i =i$. (In fact all complex numbers $(x, y)$ correspond to $x+yi$ ).

So I have just shown you how defining multiplication that way results in $i^2 = -1$.

But that multiplication isn't the multiplication I'm familiar with!, you say. Well guess what:

$a\cdot b = (a, 0) \cdot (b, 0) = (ab-0,0+0) = (ab, 0) = ab $

Yes it is!

So what I get from this is that essentially someone said: "What if there was a number that could be squared to get -1", and there you have it!

In fact, once you define addition like this:

$(a, b) + (x, y) = (a+b, x+y)$

I'm pretty sure you'll find this new system of complex numbers, $\mathbb{C}$, to be compatible with the old set of real numbers, $\mathbb{R}$ .

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    $(x,y)⋅(a,b)=(xa−yb,xb+ya)$ is just FOIL, as you're used to, observe: $(x+yi)\cdot (a+bi)$ is equivalent to the above. Just dawned on me XD, I'm a bit slow sometimes... – crackpotHouseplant Sep 20 '12 at 21:53
  • If that blows your mind, check out polar coordinates. Polar coordinates: x + yi = r ⋅ e^(Θi) (where theta is the angle from 0 in radians, and r is the distance from 0). (ed. fixed formula getting messed up in paste). –  Sep 20 '12 at 22:39

I'm surprised that, as far as I can see, no one has mentioned Paul Nahin's book "An imaginary tale : the story of √-1", pub: Princeton University Press ISBN 0-691-12798-0. It is a historical account of how √-1 became a necessary mathematical tool, and is written in an easy to read conversational style. I keep re-reading parts of it, like going over old ground again with a friend.

Two reviews give contrasting opinions: the first very favourable http://plus.maths.org/content/imaginary-tale; the second giving a long list of (alleged -- I haven't checked them independently) inaccuracies and omissions: http://www.ams.org/notices/199910/rev-blank.pdf

Harry Weston
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    The [review](http://www.ams.org/notices/199910/rev-blank.pdf) published in Notices of AMS is not very flattering. – Martin Sleziak Sep 26 '12 at 14:30
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    Thank you @Martin. I followed the link and am suitably chastened, but I did, and do, enjoy reading this book. I still think it might well be a nice introduction to complex numbers for someone who has not become acquainted with them already. – Harry Weston Sep 26 '12 at 14:50
  • The book still might be a good read (I'm not saying that the opinion of the reviewer should be taken as infallible). But link to a review might be useful for people reading your answer anyway. – Martin Sleziak Sep 26 '12 at 14:58

One answer for why imaginary (and complex) numbers are useful is that they provide solutions to polynomial equations. (The square root of -1 part comes from trying to solve the equation $x^2 = -1$, which has no real number solutions.) The Fundamental Theorem of Algebra states that any polynomial equation with real (or even complex!) coefficients has solutions in the complex number system.

The theorem doesn't always seem very powerful, because a lot of times we discard all non-real solutions. But, this isn't always the case. Linear (ordinary) differential equations can be solved by first solving an associated polynomial equation. The complex solutions to the polynomial equation end up influencing the solution to the differential equation.

Hugh Denoncourt
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  • BTW, these differential equations I speak of can be used to solve quite a few real-world problems. The type of diff-eq I mentioned arises when studying oscillators. Forced oscillators, for example, can give rise to resonance. – Hugh Denoncourt Sep 21 '12 at 04:58

This is probably not helpful for someone first learning about imaginary numbers, but my personal motivation for complex numbers is so that every linear transformation over the reals can be decomposed into a direct sum of shift plus scaling operators, ie. the Jordan normal form exists.

If you work with matrices/linear operators over the reals for long enough, this is something that "feels" like it should be true - like some sort of linear algebra version of the pigeonhole principle - but it doesn't quite work over the reals because of rotation matrices. On the other hand, rotation is "like" a scaling because if you apply a rotation twice, it's the same as rotating twice as much once, so one feels this shouldn't really be an obstruction.

In any case, complex numbers are exactly the number system you need to ensure Jordan normal form exists, where rotations are scalings of complex eigenvectors by a complex number.

Nick Alger
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@ I just want to add that generally speaking, besides the whole numbers, none of the numbers we use "exist" in the real world.

@ Sachin Kainth Hmmm. The question you ask is a deep one. The answer is far from easy. The quote above from one of the earlier answers is not right. I am not sure that "whole numbers" do "exist in the real world", let alone that "real numbers" do, or that complex numbers, quaternions or octonions don't.

The relationship between maths and the real world is extremely mysterious. It goes straight into the classic "God is a Mathematician" statement. I do not remotely have time to go into it properly here. Whole books have been written about it. Some better than others.

One viewpoint is simply to ignore questions of "reality" or relationship to the "real world" and say that complex numbers are exceedingly useful. Another approach is to go down the Clifford Algebra route, originally pioneered by William Clifford (1845-79) at my old college (Trinity, Cambridge) and which has recently seen an explosion of interest by theoretical physicists led perhaps by Stephen Gull at the Cavendish. Roger Penrose (Oxford) is also interesting on the subject of complex numbers.

But all that stuff requires some math sophistication to understand. An important prior question is "real numbers" or even fractions. There are many deeply puzzling and paradoxical questions about them. I suspect you have not been exposed to them.

Looking for someone who "totally gets it" is likely to be a vain hope. If you find them, let me know!

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Imaginary numbers can also be thought of as a simple hack mathematicians use when they want to keep units separate.

Need a result with more than one component? make it a multiple of something that won't resolve. Pretty handy.

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    That works if all you're doing is adding or subtracting, but once you start doing more complicated operations than that, the real and imaginary parts interact. I think complex numbers are best thought of as describing situations where you have two quantities that can be thought of as being separate that interact in particular ways at times, like phases. In any honest use of complex numbers, the real and imaginary parts must have the same units. – acjay Sep 21 '12 at 20:48

You can get imaginary and complex numbers with your calculator. Switch to complex mode (press the mode button and sift through the modes until you see something like CMPLX) and select it. Now input the square root of -1 again. You should get the answer i. An imaginary number is simply i with a coefficient in front, e.g. i (1i), 2i, 3i, 1/6i, 345i, sqrt3i, or whatever.

The concept is useful because you need it to pass Maths... Sometimes, for example, you may be asked to give the complex roots of a quadratic equation if the discriminant is negative.

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I just think of imaginary numbers as a definition. In the "real world" you cannot take the square root of $−1$ (which is what is happening with your calculator). However, we just define some "number", call it $i$, such that $i^2=−1$, add it to our number system and see what happens. So when you study imaginary numbers, you are just "seeing what happens".

One can then write every number as $a+ib$ where $a,b\in\mathbb{R}$ ($a$ and $b$ are real numbers) and $i^2=−1$. In his comment, ivan is taking this pair $(a,b)$ and pointing out that this pair defines a point on a plane (so, like, a piece of paper, as when you draw a graph). This is the way that people often view imaginary numbers - as points on the plane (and the plane is the Complex Plane, or an Argand diagram).

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Imaginary numbers were invented to make calculations easier. Everyone knows the quadratic formula; when Cardano was working on the formula for cubics (known as Cardano's formula), he found out that it was extremely hard to write down a formula unless you out down some symbol as a placeholder for $\sqrt{-1}$, which you manipulate like a number and which always cancelled out in the end. So he left it in. He was embarrassed by it, and called it imaginary, but the formula worked. Mathematicians later found out that imaginary numbers made a lot of formulas easier, like finding a formula for $\sin(3x)$, and so they found consistent rules for them. Ever since then, they've kept making formulas easier.

Brian Rushton
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Let's add something which hasn't quite been said yet. Suppose I take polynomials - I could be using rational coefficients, or real coefficients, for example. Because polynomials have a division algorithm, for any given polynomial $p(x)$, I can write $$p(x)=(x^2+1)q(x)+p^*(x)$$ where $p^*(x)$ is a linear polynomial, of the form $ax+b$. I want to know what happens when I replace $p(x)$ with $p^*(x)$, and whether I can do this consistently.

As an example $2x^3+4x^2+3x+1=(x^2+1)(2x+4)+x-3$

I can confirm, for example, that $$p_1(x)p_2(x)=(x^2+1)\left((x^2+1)q_1(x)q_2(x)+p^*_1(x)q_2(x)+p^*_2(x)q_1(x)\right)+p^*_1(x)p^*_2(x)$$So reducing to the linear remainder is compatible with multiplication (though the product of remainders may have to be reduced). And it is easy too to see that this works for addition. If we are not worried about multiples of $x^2+1$ we can produce a simpler polynomial arithmetic, where the results are always linear.

What use is this? Well note that $$x^2=(x^2+1)-1$$ so the linear equivalent of $x^2$ is simply $-1$ - and we have produced a simplified form of polynomial arithmetic in which it makes sense to say $x^2=-1$. Of course we haven't said what $x$ is at any stage, so we haven't yet said it is a number of any kind. We can, however, call any multiple of $x$ in this system an "imaginary number" if we so wish.

The geometric explanations are definitely the ones to go for first. But the more abstract algebraic approach becomes very significant in more advanced work. And the abstract approach shows that we can calculate in this way if it happens to be useful - we have a consistent system. And it has proven to be very useful indeed, not least because (this is a geometric insight) it enables us to treat the points on the unit circle as numbers - an insight which transforms the way in which we think about periodic functions and waves.

Mark Bennet
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It's an extension field . . . but since you probably don't know that, the terminology is horrible! Just think of imaginary numbers as the completion of real numbers so that you can find solutions to the equation

$x^2 + 1 = 0.$

If you set $i = \sqrt{-1}$, then $i$ and $-i$ are solutions to this polynomial. There are no 'real solutions'.

It is known that any univariate system of degree $n$ has exactly $n$ solutions in the complex plane. This can be generalized further to multivariate square systems in that the max number of solutions is the multiplication of the largest degree of each function.

For example,

$x_1^3 + x_1*x_2 + 4 = 0$ $x_1^2*x_2 + x_1 + x_2 - 2 = 0$

has (at most) $3*4 = 12$. This is known as Bezout's Theorem and is a result of classical algebraic geometry. Letting $i = \sqrt{-1}$ is necessary to find all the solutions (and it is possible to find these solutions numerically, up to an arbitrary choice of precision).

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As no one has mentioned about this, here is an excellent intuitive video series (that shouldn't be missed) by Welch Labs as Imaginary Numbers are Real.

These are a set of 13 videos that beautifully help with the visualization of complex numbers from A Simple Complex Plane, all the way up to Riemann Surfaces. Here is something from the videos to get you interested.

Consider a Function $f(x)=x^{2}+1$.

Here is the plot of the function in the real x-y plane:


(Source: WolframAlpha)

Now according to the Fundamental Theorem of Algebra we should have n-roots for n-th degree polynomial but if you consider the graph for the given function it doesn't appear to intersect the x-axis right ?

Well, the thing is, we are not seeing it correctly and have not included a fundamental set of numbers : Complex Numbers which have both real and imaginary part but don't get confused yet as both the parts are quite real.The below GIF from the above videos beautifully plots the the function in the Complex plane (The vertical axis that comes out of the paper being the imaginary axis).

complex plot of function by welch labs

(Source: Welch Labs)

So you can see that the function actually intersects the x-axis but in a different dimension and that is the imaginary plane. This gives an excellent intuitive sense of why imaginary numbers are very real! I would recommend the users to view all the videos as there's a lot of graphic animations for better visualization and understanding.

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'What are imaginary numbers?'

The most honest answer I can think of is: we don't know.

But they are a very, very useful tool, as with many other mathematical abstractions that we really don't fully philosophically understand, and yet they are the true foundations of our modern civilizations, because they are practical and they proven to work very well.

A simple reasoning (I often give to my kid) goes as follows: what is the number that multiplied by itself gives 4? The answer is 2, or minus 2, of course. Then, what is the number that multiplied by itself gives 1? The answer is 1, or minus 1.

Now, what is the number that multiplied by itself gives 2? The answer for this one is not so simple (if you are not allowed to use ellipsis), its $\sqrt{2}$, or $-\sqrt{2}$. We have to start using more complicated symbols for this one now. Square root of two is not the most irrational number we can think of, as with $\pi$ for example, but it is so difficult to declare it on a finite character sequence basis that we have to resort to a simpler, and yet more obscure, symbol or set of symbols.

But, now, what is the number that multiplied by itself gives $-1$? We don't know! And perhaps we will never know. So, we decide to encapsulate all of our ignorance by naming it $i$, and it all works very well by only dealing with this symbol. It is an unknown, but it serves us very duly.

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