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

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

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    $\begingroup$ 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. $\endgroup$ Aug 25, 2010 at 23:57
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    $\begingroup$ 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. $\endgroup$ Aug 26, 2010 at 0:15
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    $\begingroup$ Do you want an explanation just using algebraic properties of complex numbers? $\endgroup$ Aug 27, 2010 at 16:54
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    $\begingroup$ 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. $\endgroup$ Aug 27, 2010 at 19:49
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    $\begingroup$ 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. $\endgroup$ Dec 11, 2016 at 0:21

10 Answers 10


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

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

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    $\begingroup$ An equation might have two solutions, but an expression (like $\sqrt{i}$) can only have one value. $\endgroup$ Mar 1, 2019 at 22:39

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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    $\begingroup$ WHat did you use to draw that? $\endgroup$
    – claws
    Oct 20, 2010 at 11:17
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    $\begingroup$ geogebra: www.geogebra.org $\endgroup$
    – zar
    Oct 20, 2010 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, $-e^{\pi i/4}=e^{-3\pi i/4}$

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    $\begingroup$ 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.) $\endgroup$
    – Matt E
    Aug 26, 2010 at 1:17
  • $\begingroup$ 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.) $\endgroup$
    – Matt E
    Aug 26, 2010 at 1:22
  • $\begingroup$ Yes, thanks for correcting Matt, I shall fix the answer now. $\endgroup$ Aug 26, 2010 at 1:32
  • $\begingroup$ 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 $\endgroup$ Aug 26, 2010 at 11:50
  • $\begingroup$ Dear J., You're welcome! $\endgroup$
    – Matt E
    Aug 27, 2010 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.

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    $\begingroup$ +1 for algebraic closure! As you rightly point out it is what makes complex numbers so important. $\endgroup$
    – user892
    Aug 31, 2010 at 0:22
  • $\begingroup$ So would the next higher imaginary number be j? $\endgroup$ Sep 30, 2010 at 0:14
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    $\begingroup$ @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. $\endgroup$ Sep 30, 2010 at 1:46
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    $\begingroup$ 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.) $\endgroup$ Sep 30, 2010 at 1:51
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    $\begingroup$ 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? $\endgroup$ Mar 1, 2019 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,

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

  • $\begingroup$ @Mangaldan. I can't help it. :-) $\endgroup$ Aug 26, 2010 at 13:28
  • $\begingroup$ 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). $\endgroup$
    – user892
    Aug 31, 2010 at 0:23
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    $\begingroup$ 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). $\endgroup$ Mar 1, 2019 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 $x=\pm\frac{\sqrt{2}}{2}(1+i)$.

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    $\begingroup$ So, which one of them is $\sqrt{i}$? $\endgroup$ Mar 1, 2019 at 23:03
  • $\begingroup$ @WolfgangKais $(1.4142...)^2=(-1.4142...)^2=2$, so which one of them is $\sqrt{2}$? :) $\endgroup$
    – George K.
    Jul 6, 2023 at 8:17
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    $\begingroup$ @GeorgeK. The suaqre root of a (non-negative) real number is non-negative by definition, but is there a similar decision for "the" square root of other (complex) numbers? $\endgroup$ Jul 8, 2023 at 17:59
  • $\begingroup$ @WolfgangKais In the reals sometimes it's conventional to use $\sqrt{\cdot}$ to indicate the non-negative square root. The complex analogue is the Principle Value, in this case of the complex logarithm, which is actually hiding in this question: $i^{\frac{1}{2}} = \exp\left(\frac{1}{2}\log{i}\right) = \exp\left(\frac{1}{2}(\frac{\pi}{2}+2n\pi)i\right) = \pm \frac{1}{\sqrt{2}}(1 + i)$, depending on the parity of $n$. The principle value corresponds to $n=0$, so we might consider the $+$ case as having some kind of conventional priority. $\endgroup$
    – George K.
    Jul 11, 2023 at 12:02


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


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



  • $\begingroup$ Much better explained, four years ago, in several answers. $\endgroup$
    – Did
    May 12, 2015 at 17:53
  • $\begingroup$ @Did who? Please share a link. $\endgroup$ Mar 1, 2019 at 23:10
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    $\begingroup$ @Did no need to be rude to a new person $\endgroup$ Nov 5, 2021 at 21:28

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)

  • $\begingroup$ Sorry, don't know yet how to properly enter mathematical symbols and constants. $\endgroup$
    – stevenvh
    Aug 26, 2010 at 15:56
  • $\begingroup$ 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. $\endgroup$ Aug 27, 2010 at 3:09
  • $\begingroup$ +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. $\endgroup$
    – Tyler
    Aug 27, 2010 at 6:24
  • $\begingroup$ MatrixFrog: "Argand plane" is another term used for the complex plane. $\endgroup$ Aug 27, 2010 at 11:01
  • $\begingroup$ @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? $\endgroup$
    – stevenvh
    Aug 27, 2010 at 16:41

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


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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