Take the 2-minute tour ×

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

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

share|improve this question
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
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 0:15
Do you want an explanation just using algebraic properties of complex numbers? –  Américo Tavares Aug 27 '10 at 16:54
This is a fairly common and natural question, you should lookup the similar threads math.stackexchange.com/questions/2914/… math.stackexchange.com/questions/3210/… math.stackexchange.com/questions/1211/… etc. –  AD. Aug 27 '10 at 19:49

9 Answers 9

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

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

share|improve this answer

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

share|improve this answer
WHat did you use to draw that? –  claws Oct 20 '10 at 11:17
geogebra: www.geogebra.org –  zar Oct 20 '10 at 18:09

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

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

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

share|improve this answer
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 1:17
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 1:22
Yes, thanks for correcting Matt, I shall fix the answer now. –  Ｊ. Ｍ. Aug 26 '10 at 1:32
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. Aug 26 '10 at 11:50
Dear J., You're welcome! –  Matt E Aug 27 '10 at 0:52

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.

share|improve this answer
+1 for algebraic closure! As you rightly point out it is what makes complex numbers so important. –  Ninefingers Aug 31 '10 at 0:22
So would the next higher imaginary number be j? –  Talvi Watia Sep 30 '10 at 0:14
@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 1:46
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 1:51

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

$$ \sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{0_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 $$

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,

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

share|improve this answer
+1 for generality. :D –  Ｊ. Ｍ. Aug 26 '10 at 13:10
@Mangaldan. I can't help it. :-) –  a.r. 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). –  Ninefingers Aug 31 '10 at 0:23

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

share|improve this answer

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)

share|improve this answer
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. –  a.r. Aug 27 '10 at 3:09
+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. –  MatrixFrog Aug 27 '10 at 6:24
MatrixFrog: "Argand plane" is another term used for the complex plane. –  Ｊ. Ｍ. 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

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

share|improve this answer

You can use De Moivre's Theorem. First note that $i$ can be expressed in the form $ 1 \, \text{cis} \left( \frac{\pi}{2} \right) $. Hence, $$ \sqrt{1} = \left( 1 \, \text{cis} \left( \frac{\pi}{2} \right)\right)^{\tfrac{1}{2}} = \pm \left[ 1 \, \text{cis}\left(\frac{\pi}{4}\right) \right] = \pm \left( \frac{1}{\sqrt{2}} + \frac {i}{\sqrt{2}} \right). $$

share|improve this answer

Your Answer


