If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?

Is it used or considered often in mathematics? How is it notated?

Martin Sleziak
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Gordon Gustafson
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    The square root of i is (1 + i)/sqrt(2). [Try it out my multiplying it by itself.] It has no special notation beyond other complex numbers; in my discipline, at least, it comes up about half as often as the square root of 2 does --- that is, it isn't rare, but it arises only because of our prejudice for things which can be expressed using small integers. – Niel de Beaudrap Aug 25 '10 at 23:57
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    Do you know the exponential notation for complex numbers, i.e. $e^{ix}=\cos x+i\sin x$?In my unser bellow I assumed you would. – Américo Tavares Aug 26 '10 at 00:15
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    Do you want an explanation just using algebraic properties of complex numbers? – Américo Tavares Aug 27 '10 at 16:54
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    This is a fairly common and natural question, you should lookup the similar threads http://math.stackexchange.com/questions/2914/how-to-combine-complex-powers http://math.stackexchange.com/questions/3210/simple-complex-number-problem-1-1 http://math.stackexchange.com/questions/1211/non-integer-powers-of-negative-numbers etc. – AD - Stop Putin - Aug 27 '10 at 19:49
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    Just adding, complex numbers exhibit a property known as closure; any operations on them involving +, −, ×, ÷, $\sqrt{\cdots}$, $\sqrt[n]{\cdots}$, etc. will always produce an answer that is also complex. – DividedByZero Dec 11 '16 at 00:21
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    Here's how to find the square root(s) of any complex number $z$, not just $z=i$: https://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – Hans Lundmark Nov 12 '20 at 18:55

10 Answers10


Let $z=(a+bi)$ be a complex number which is a square root of $i$, that is $$i=z^2=(a^2-b^2)+2abi.$$ Equating real and imaginary parts we have, $$a^2-b^2=0, 2ab=1$$

The two real solutions to this pair of equations are $a={1 \over \sqrt{2}},b={1 \over \sqrt{2}}$ and $a=-{1 \over \sqrt{2}},b=-{1 \over \sqrt{2}}$. The two square roots of $i$ therefore are

$$\pm {1 \over \sqrt{2}} (1+i)$$

Jyotirmoy Bhattacharya
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Visually, the square root of a complex number (written in polar coordinates) $(\rho,\theta)$ is the number $(\sqrt\rho,\theta/2)$.alt text

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$i^\frac1{2}=\left(e^{\pi i/2}\right)^\frac1{2}=e^{\pi i/4}$

$e^{\pi i/4}=\cos\left(\frac{\pi}{4}\right)+i \sin\left(\frac{\pi}{4}\right)$

or simplified,


This is of course the "principal value"; the other value (thanks Matt E!) is the "negative square root", $-\frac{1+i}{\sqrt{2}}$ or in exponential form, $-e^{\pi i/4}=e^{-3\pi i/4}$

J. M. ain't a mathematician
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    There are only two square roots of ii (as there are two square roots of any non-zero complex number), namely $\pm(1+i)/\sqrt{2}$. In the context of your answer, what happens is that the different values are $e^{(\pi i/2+2\pi ik)/2} = e^{\pi i/4 + \pi i k}$; but the value of this depends only on the parity of $k$, and so gives just two values, namely $\pm e^{\pi i/4} = \pm (1+ i)/\sqrt{2}$. Also, you have a typo, or a misplaced expression: when you write the other values, you want $\pi i k$, not $2\pi i k.$ (Adding a multiple of $2 \pi i$ doesn't change anything.) – Matt E Aug 26 '10 at 01:17
  • Correction to the above comment: "two square roots of ii" should read "two square roots of $i$". (I couldn't face editing it a second time and dealing with all the messed up formulas.) – Matt E Aug 26 '10 at 01:22
  • Yes, thanks for correcting Matt, I shall fix the answer now. – J. M. ain't a mathematician Aug 26 '10 at 01:32
  • Matt E: I think I now remember why I made that mistake in the first version of this answer; I had somehow conflated the ways to show the explicit form of $\sqrt{i}$ and $i^i$. :D – J. M. ain't a mathematician Aug 26 '10 at 11:50
  • Dear J., You're welcome! – Matt E Aug 27 '10 at 00:52
  • @MattE As "square root" is a function, there can't be 2 square roots for just one argument. There are 2 solutions for an equation like $x^2=a$ (if $a \ne 0$), yes, and these are $\sqrt{a}$ and $-\sqrt{a}$, but there can only be one $\sqrt{a}$. – Wolfgang Kais Mar 01 '19 at 22:21

As you can see, there are several ways to find the answer to your question, from jmoy's straightforward algebra calculation, to using Euler's formula $e^{r+i\theta} = e^r(\cos(\theta)+i \sin(\theta))$, to geometric interpretations of complex numbers.

One fact that hasn't been mentioned yet is that the complex numbers are algebraically closed. This means that every algebraic equation using complex numbers has all of its solutions in the complex numbers. So, the equation $x^2=i$ has both of its solutions in the complex numbers; and the equations $x^4 = -7-12i$ and $x^4+(\pi -8i)x^3+x-\sqrt{5}=0$ each have all four of their solutions in the complex numbers.

The real numbers are not algebraically closed: the equations $x^2=-1$ and $x^4+7x^2+\pi=0$ cannot be solved unless we use complex numbers.

This is one of the main reasons complex numbers are so important; they are the algebraic closure of the real numbers. You will never need "higher levels" of imaginary numbers or new mysterious square roots; numbers of the form $a+bi$ are all you need to find any root of real or complex polynomials.

Jonas Kibelbek
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    +1 for algebraic closure! As you rightly point out it is what makes complex numbers so important. –  Aug 31 '10 at 00:22
  • So would the next higher imaginary number be `j`? – Talvi Watia Sep 30 '10 at 00:14
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    @Talvi Watia. No, there is no "higher imaginary number" because we don't need them; all we need is "i". We need "i" to solve equations like "x^2 = -1"; but as the other answers show, we can solve "x^2 = i" without any new numbers. The point of my answer is that we can solve *any* polynomial now. You will never find a polynomial that requires new kinds of numbers to solve. – Jonas Kibelbek Sep 30 '10 at 01:46
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    P.S. Engineers often write "j" instead of "i" for the square root of -1. It's not a different number, just different notation. There is also an algebra called the Quaternions that uses "i", "j", and "k" as kinds of imaginary numbers, but it's a different sort of object than the complex numbers. In the Quaternions, some of the familiar rules of multiplication no longer work. (Quaternion multiplication is not commutative.) – Jonas Kibelbek Sep 30 '10 at 01:51
  • As $\{\} \subset \mathbb R$, all real solutions of the real algebraic equation $x^2=-1$ are real numbers. How can anyone know whether an equation has solutions or not without knowing the "feasible set" to search in? Maybe $1=2$ has infinitely many solutions in some superset of $\mathbb C$, so maybe the complex numbers aren't **algebraically closed**? – Wolfgang Kais Mar 01 '19 at 22:56

More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that

$$ z^n = z_0 \ , \qquad \qquad \qquad \qquad [1] $$

you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes

$$ \left( r e^{i \theta}\right)^n = r_0 e^{i\theta} \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0} \ . $$

Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:

$$ r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0} $$

since $r, r_0 \geq 0$.

As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:

$$ n\theta = \theta_0 + 2k\pi \qquad \Longleftrightarrow \qquad \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots $$

It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally

$$ \sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{\theta_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 $$

are all the complex $n$-th roots of $z_0$.


(1) For $n=2$, we obtain that every complex number has exactly two square roots:

$$ \begin{align} \sqrt{z_0} &= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1 \\\ &= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)} \ . \end{align} $$

For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain

$$ \sqrt{i} = \begin{cases} e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\\ e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \ . \end{cases} $$

Also, if $z_0 = -1 = e^{i\pi}$,

$$ \sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ . $$

(2) For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th roots of unity:

$$ \sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ . $$

For instance, if $n= 2$, we get

$$ \sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad e^{i\pi}= -1 $$

and for $n= 4$,

$$ \sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ , $$

that is,

$$ \sqrt[4]{1} = 1, i, -1 , -i \ . $$

Agustí Roig
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  • @Mangaldan. I can't help it. :-) – Agustí Roig Aug 26 '10 at 13:28
  • Studied as an A-level/first year undergrad topic in the UK you might find this called "roots of unity", the approach to which leads you to understand sqrt(i) is a special case of sqrt(z \in C). –  Aug 31 '10 at 00:23
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    We must not mix up $n$-th root (which is an expressionthat can only take one value) and $n$-th roots (which is a set of solutions of an equation). – Wolfgang Kais Mar 01 '19 at 23:00

With a little bit of manipulation you can make use of the quadratic equation since you are really looking for the solutions of $x^2 - i = 0$, unfortunately if you apply the quadratic formula directly you gain nothing new, but...

Since $i^2 = -1$ multiply both sides of our original equation by $i$ and you will have $ix^2 +1 =0$, now both equations have exactly the same roots, and so will their sum. $$(1+i)x^2 + (1-i) = 0 $$
Aplly the quadratic formula to this last equation and simplify an you will get $x=\pm\frac{\sqrt{2}}{2}(1+i)$.


If you understand Argand diagrams (the representation of complex numbers in the complex plane) and can envision the unit circle in it, you can easily do this in your head:
-1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated over half the angle, so $i$, or $\sqrt{-1}$ is 1 rotated over $\pi/2$ radians. To find $\sqrt{i}$ you just half the angle again: $\pi/4$ radians. The corresponding real and imaginary parts are $\cos\frac{\pi}{4}$ and $\sin\frac{\pi}{4}$ resp.

LaTexified (thanks Agosti)

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  • Sorry, don't know yet how to properly enter mathematical symbols and constants. – stevenvh Aug 26 '10 at 15:56
  • If you "click" the link at the right hand side of the "edit" at the end of some messages, you'll get a quick introduction to LaTeX. – Agustí Roig Aug 27 '10 at 03:09
  • +1 for introducing me to the term "Argand diagrams" which I'd never heard before. I always just heard "representations in the complex plane" or something. – Tyler Aug 27 '10 at 06:24
  • MatrixFrog: "Argand plane" is another term used for the complex plane. – J. M. ain't a mathematician Aug 27 '10 at 11:01
  • @Agusti - I don't see this link to the intro to LaTex, but I can find this elsewhere on the 'Net. I tried replacing "pi" with "\pi" but the symbol doesn't appear. Is there something else I should know? – stevenvh Aug 27 '10 at 16:41
  • @stevenvh. No: there isn't any introduction to LaTeX, but LateX code itself, I meant. See Américo's answer up here? It says "edited yesterday" in the end, doesn't it? Click on "yesterday". As for \pi , you should write it with two dollars around (one before, one behind). Ah, writting questions or answers with LaTeX, you should wait five seconds without typing anything to see what happens. – Agustí Roig Aug 27 '10 at 18:34
  • @stevenh. I forgot: after clicking on the word on the right of "edited" you must click on "view source". – Agustí Roig Aug 27 '10 at 19:26
  • stevenvh: As an example of what Agusti's talking about: `$\pi$` gives $\pi$. – J. M. ain't a mathematician Aug 27 '10 at 23:18


First we look to $\left|\sqrt{i}\right|$:

$$\left|\sqrt{i}\right|=\left|\sqrt{\frac{1}{2}+i-\frac{1}{2}}\right|=\left|\sqrt{\frac{1+(0+2i)-1}{2}}\right|=\left|\sqrt{\frac{1+2(0+1i)+(0+1i)^2}{2}}\right|=$$ $$\left|\sqrt{\frac{(1+(0+1i))^2}{2}}\right|=\left|\frac{\sqrt{(1+(0+1i))^2}}{\sqrt{2}}\right|=\left|\frac{1+(0+1i)}{\sqrt{2}}\right|=$$ $$\left|\frac{(1+(0+1i))\sqrt{2}}{2}\right|=\frac{\left|(1+(0+1i))\sqrt{2}\right|}{|2|}=\frac{\sqrt{2}|1+(0+1i)|}{2}=$$ $$\frac{|1+1i|}{2}=\frac{\sqrt{2}\sqrt{1^2+1^2}}{2}=\frac{\sqrt{2}\sqrt{2}}{2}=\frac{2}{2}=1$$

Now the argument of $\sqrt{i}$:

It's positive so on the complex axces, so $\sqrt{\sqrt{-1}}$ gives us $1e^{\frac{1}{4}\pi i}$ so the argument of $\sqrt{i}$ is $\frac{1}{4}\pi$


$$\sqrt{i}=\left|\sqrt{i}\right|e^{\arg\left(\sqrt{i}\right)i}=1e^{\frac{1}{4}\pi i}=1\left(\cos\left(\frac{1}{4}\pi\right)+\sin\left(\frac{1}{4}\pi\right)i\right)=$$ $$\cos\left(\frac{1}{4}\pi\right)+\sin\left(\frac{1}{4}\pi\right)i=$$ $$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$



Jan Eerland
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Gives a nice concise overview: http://www.wolframalpha.com/input/?i=Sqrt(i)

Lie Ryan
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You can use De Moivre's Theorem. At first note that $i$ can be expressed in the form: $ 1 \, \text{cis} \left( \frac{\pi}{2} \right) $.

Hence, $$ \sqrt{i} = \left( 1 \, \text{cis} \left( \frac{\pi}{2} \right)\right)^{\tfrac{1}{2}} = \pm \left[ 1 \, \text{cis}\left(\frac{\pi}{4}\right) \right] = \pm \left( \frac{1}{\sqrt{2}} + \frac {i}{\sqrt{2}} \right). $$

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Ahaan S. Rungta
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