Wolfram says, there are only two roots, but $\sqrt{i}$ already gives two roots. So if we express them in Cartesian form we can take square roots of them separately and end up with four roots.

$$\sqrt{i}=e^{\frac{i\pi}{4}}=\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$$ But also $$\sqrt{i}=e^{\frac{-3i\pi}{4}}=-\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}$$ Then take square roots of each of those and you end up with $$\sqrt{\sqrt{i}}=e^{\frac{i\pi}{8}},e^{\frac{5i\pi}{4}},e^{\frac{-3i\pi}{4}},e^{\frac{-7i\pi}{4}}$$

It works, doesn't it? Raise each of those to the power of 4 and you get $i$. What am I missing here?

  • $\begingroup$ What is the exact expression that you're plugging into Wolfram? $\endgroup$ Commented Feb 3, 2016 at 23:11
  • $\begingroup$ (sqrt(sqrt(i). it recognizes it correctly $\endgroup$ Commented Feb 3, 2016 at 23:12
  • $\begingroup$ Is $e^{\frac{5\pi i}{4}}$ and $e^{-\frac{3\pi i}{4}}$ really different numbers? Same with the other two. Try writing your numbers with the argument in $[0,2\pi)$. $\endgroup$
    – Winther
    Commented Feb 3, 2016 at 23:15
  • $\begingroup$ $\sqrt{\sqrt{i}}$ is single-valued (and takes value $e^{i\pi/8}$) under usual definition of square root, because we are taking a particular branch so that $z \mapsto \sqrt{z}$ is a single-valued function. If you are discussing the number of solutions of $z^4 = i$, then it has exactly 4 solution, all of which you have listed. $\endgroup$ Commented Feb 3, 2016 at 23:18
  • $\begingroup$ Thank You all, it's me being silly rather than anything as Winther suggested two roots are repeated. So the polynomial will have 4 roots but only two of them are distinct. That also explains why Wolfram only bother with listing two of them. Thanks for help $\endgroup$ Commented Feb 3, 2016 at 23:20

4 Answers 4


Your list of four solutions only has two - as has been pointed out you listed two of them twice. (In the original post, anyway...) But there are four roots.

First, $i=e^{i\pi/2}=e^{i5\pi/2}$ leads gives two values for $\sqrt i$, namely $e^{i\pi/4}$ and $e^{i5\pi/4}$. Each of these has two square roots: Since $e^{i\pi/4}=e^{i9\pi/4}$ it has the two square roots $e^{i\pi/8}$ and $e^{i9\pi/8}$. Similarly $e^{i5\pi/4}=e^{i13\pi/4}$ has the two square roots $e^{i5\pi/8}$ and $e^{i13\pi/8}$. For a total of four values of $\sqrt{\sqrt i}$, namely $$e^{i\pi/8},e^{i5\pi/8},e^{i9\pi/8},e^{i13\pi/8}.$$

NOTE Others have suggested that although a complex number has two square roots, the notation $\sqrt z$ refers to only one of them. I think it's wrong to put it that way; that notation refers to only one of them if we have clearly stated in advance which "branch" we're referring to. But Wolfram is wrong in any case. By my lights, $i$ has two square roots, each of which has two square roots, for a total of four as above. But if we are saying that the notation $\sqrt z$ refers to one square root, then Wolfram should say there's only one value for $\sqrt{\sqrt i}$, namely the one square root of the one square root.

I think four is the right number. With that convention there's only one; there's no sensible way to give exactly two values, as WA does. You can't believe everything they tell you.

  • $\begingroup$ Thank You, That cleared everything. I wouldn't start with expressing i as two "different" exponential. I guess maybe such reasoning will come with experience. :) $\endgroup$ Commented Feb 3, 2016 at 23:44
  • $\begingroup$ So a complex number has four fourth roots. You can find the fourth roots of $i$ by writing $i$ as an exponential in four different ways. There are actually infinitely many ways to write $i$ as an exponential, but it happens they lead to only four fourth roots. So what you should do next is find all the infinitely many values of $i^i$, obtained from the infinite list of ways to wrrite $i=e^{it}$. $\endgroup$ Commented Feb 4, 2016 at 0:00
  • $\begingroup$ I must disagree with this. The square root function is perfectly well defined as a function. Have a look at my answer if you'd like to understand why Alpha responds the way that it does. $\endgroup$ Commented Feb 4, 2016 at 13:55
  • $\begingroup$ @MarkMcClure The principal branch of the square root is what's perfectly well defined as a function. You can say that's "the square root" if you want but that's in no way a standard definition. $\endgroup$ Commented Feb 4, 2016 at 14:23
  • $\begingroup$ @DavidC.Ullrich As I said in my response to this same comment on my answer, the question concerns WolframAlpha's response, which is built on top of Mathematica's Sqrt function which is well known to return the principal square root. $\endgroup$ Commented Feb 4, 2016 at 14:26

Trick is to introduce an unknown, $z$. Write $z=\sqrt{ \sqrt{i}}$. We want to know how many values are possible for $z$. Squaring we get $z^2=\surd i$. Again squaring we get $z^4-i=0$. This is a polynomial of degree 4 over the complex numbers. So it has 4 roots. Taking the derivative shows that the roots are distinct.

SO we have 4 distinct values for the number.

EDIT (to clarify the doubts raised by SHailesh and Piquito in the comments):

For a polynomial $f(z)$ to have $\alpha$ as repeated root means it is of the form $(z-\alpha)^2 g(z)$ for some other polynomial $g(z)$.Calculating derivative of $f(z)$ using product rule makes it clear that $\alpha$ will also be a root of the derivative $f'(z)$; and converse is also true.

@Piquito: Fundamental theorem of algebra states that the number of roots is equal to the degree of the polynomial only when we count them with multiplicity and include complex roots, real irrational roots.

  • $\begingroup$ I did not understand taking the derivative part. Could you elaboratte ? $\endgroup$
    – Shailesh
    Commented Feb 4, 2016 at 0:26
  • $\begingroup$ Why over $\mathbb C$ and not over $\mathbb Q$ so you have eight values? $\endgroup$
    – Piquito
    Commented Feb 4, 2016 at 0:26
  • 1
    $\begingroup$ @Shailesh and Piquito: My answer above has been edited and revised. $\endgroup$ Commented Feb 4, 2016 at 2:38

As has been pointed out in the other answers, there are four complex numbers $z$ with the property that $z^4=i$. If you would like to understand the behavior exhibited by WolframAlpha, however, you should consider the fact that its square root function is (in fact) a function built on top of Mathematica's Sqrt command. It returns exactly one number, namely $$\sqrt{r e^{i\theta}} = \sqrt{r} e^{i\theta/2},$$ when $r\geq0$ and $-\pi<\theta\leq\pi.$ As a result, $\sqrt{\sqrt{i}}$ is uniquely determined to be $$\sqrt{\sqrt{i}} = \cos \left(\frac{\pi }{8}\right)+i \sin \left(\frac{\pi }{8}\right) \approx 0.92388 + 0.382683 i.$$ That is exactly what Wolfram Alpha has told you in the first couple of pods.

When you say that Alpha says there are two roots, I assume that you are referring to the last couple of pods that refer to "all 2nd roots of $(-1)^{1/4}$". First off, those pods would be considered secondary. They arise because WolframAlpha typically produces secondary pods containing related information in an effort to respond to the various things that a human user might have meant in their input.

The reason that there are two is because sqrt(sqrt(i)) essentially parses as

sqrt(a complex number)

so you get the two roots of that complex number in the secondary pod. You would get the same result, if you type sqrt((1+i)/sqrt(2)) because $$\sqrt{i} = (1+i)/\sqrt{2}.$$

Finally, if you want all four fourth roots of $i$, simply type

all fourth roots of i

You can get more information on how WolframAlpha deals with roots in this blog post.

  • $\begingroup$ Most people would say that's not "the square root", rather it's the principal branch of the square root. $\endgroup$ Commented Feb 4, 2016 at 14:21
  • $\begingroup$ @DavidC.Ullrich Exactly - and that's what Mathematica's Sqrt function has returned since 1988. $\endgroup$ Commented Feb 4, 2016 at 14:24
  • $\begingroup$ So? Are you claiming that Mathematica is the universal authority on mathematical definitions? I'm not disputing that that's what the Mathematica Sqrt function returns - that doesn't make it The square root in mathematics. $\endgroup$ Commented Feb 4, 2016 at 14:27
  • $\begingroup$ @DavidC.Ullrich No, I am not claiming that Mathematica is the universal authority. I am answering a question about the behavior of a particular software tool, however, so I think that a proper understanding of that behavior requires an understanding of the conventions employed by that software. $\endgroup$ Commented Feb 4, 2016 at 14:29
  • $\begingroup$ If you say so. It's not clear to me that the question is about the behavior of Mathematica, as opposed to about what the mathematical truth is. There are only two question marks in the original post - those questions appear to me to be questions about the validity of the OP's calculations. $\endgroup$ Commented Feb 4, 2016 at 14:49

We can write $i$ as $i=e^{i(\pi/2+2k\pi)}$ for any integer $k$. Then, taking a square root, we see that

$$\sqrt{i}=e^{i(\pi/4+k\pi)} \tag 1$$

for any integer $k$. Note that there are two distinct values of $\sqrt i$ in $(1)$; for $k=-1,0$, we have $\sqrt i = e^{-3i\pi/4}$ and $\sqrt i = e^{i\pi/4}, respectively.

Taking the square root of the right-hand side of $(1)$ gives


for which there are $4$ distinct values corresponding to $k=-2,-1,0,1$. The solutions are $e^{-i7\pi/8}$, $ e^{-i3\pi/8}$, $e^{i\pi/8}$, and $e^{i5\pi/8}$


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