What is $\sqrt{i}$?

Question

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

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

Answer 1

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

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

Answer 2

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

Answer 3

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

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

or simplified,

$\frac{1+i}{\sqrt{2}}$

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, $-e^{\pi i/4}=e^{-3\pi i/4}$

Answer 4

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.

Answer 5

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 $z_0$.

Examples

(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,

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

Answer 6

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 $x=\pm\frac{\sqrt{2}}{2}(1+i)$.

Answer 7

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.

edit
LaTexified (thanks Agosti)

Answer 8

$$\sqrt{i}=\left|\sqrt{i}\right|e^{\arg\left(\sqrt{i}\right)i}$$

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 $\frac{1}{4}\pi$

$-------$

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

So:

$$\sqrt{i}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$

Answer 9

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

Answer 10

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