It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as $i$ is not a real number. Can anyone help?

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    $\begingroup$ It depends on which branch cut of $\sqrt{\cdot}$ you take, but under the principal branch cut $(-\infty, 0]$ and with the recursive formula $z_0 = i$, $z_{i+1} = i\sqrt{z_n}$ we have $\lim z_n = -1$. $\endgroup$ Mar 1, 2018 at 21:15
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    $\begingroup$ The two highest-voted answers (as of writing) are incorrect because they do not show that the expression even has a limit, only that if the limit exists then it is $-1$. $\endgroup$
    – user856
    Mar 2, 2018 at 4:31
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    $\begingroup$ How is the first claim obvious? Why do you think the answer "might be $-1$"? What's up with the title? $\endgroup$
    – JiK
    Mar 2, 2018 at 6:16
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    $\begingroup$ @JiK is right: this question needs to be retooled from soup to nuts. This isn't comp.ai.philosophy. $\endgroup$
    – neuronet
    Mar 2, 2018 at 13:53
  • $\begingroup$ @Rahul does my answer show that, or is something still missing? $\endgroup$
    – LLlAMnYP
    Mar 2, 2018 at 14:28

7 Answers 7


\begin{eqnarray*} x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\ x=a^{ 1+1/2+1/4+1/8+\cdots} \\ x=a^2 \end{eqnarray*} So it would seem that \begin{eqnarray*} i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}. \end{eqnarray*}

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    $\begingroup$ That's got to be the best imaginary root series I have ever seen. - So it would seem. $\endgroup$ Mar 2, 2018 at 2:49
  • $\begingroup$ @ClassicEndingMusic You sure you mean series and not infinite product? $\endgroup$
    – BCLC
    Mar 15, 2018 at 8:04

I don't know if it's absolutely correct, but I am posting it.

If we write $i $ as $e^{i\pi/2} $, then the given series becomes:

\begin{align} & e^{i\pi/2} \sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}} \cdots}}} \\[8pt] = {} & e^{i\pi \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \cdots \right)} \\[8pt] = {} & e^{i\pi \left( \frac{1/2}{1-1/2} \right)} \\[8pt] = {} &\boxed{e^{i\pi}=-1} \end{align}

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    $\begingroup$ @IwillnotexistIdonotexist But you don't even need it as seen in Michael's answer. $\endgroup$
    – M. Winter
    Mar 2, 2018 at 13:17
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    $\begingroup$ You never even use the $e^{i\pi/2}$ representation, you could just as well have used $i$ and $i^2$ instead. $\endgroup$ Mar 2, 2018 at 16:08
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    $\begingroup$ Why does this have so many upvotes? The transformation to/from circular coordinates is completely superfluous, it serves no purpose. $\endgroup$ Mar 2, 2018 at 19:11
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    $\begingroup$ @BlueRaja-DannyPflughoeft On the contrary, I think it serves a very important purpose for the OP, who is uncertain about the relationship between real and complex numbers. A proof that exploits the key relationship between them (Euler's formula) is therefore essential from an algebra point of view. But what's more, this answer exposes a lovely geometric interpretation: The composition of rotations by 1/4, 1/8, 1/16... turns around the circle is 1/2 turns (a 180 degrees turn, a negation). Many other answers instead simply affirm the OP's own answer while relying on knowledge unavailable to OP. $\endgroup$ Mar 2, 2018 at 19:28

Let $z = r e^{i\theta} \in \mathbb{C}$ and $(z_n)_{n \geq 0}$ be defined by

$$ z_0 = z, \qquad z_{n+1} = z \sqrt{z_n} $$

where $\sqrt{\cdot} = \exp(\frac{1}{2}\log(\cdot))$ is the principal square root. In particular, if we define $m : \mathbb{R} \to \mathbb{R}$ by

$$ m(x) = \begin{cases} x, & \text{if } x \in (-\pi, \pi] \\ m(x + 2\pi) & \text{for all } x \in \mathbb{R} \end{cases} $$

then it follows that $\sqrt{re^{i\theta}} = \sqrt{r}e^{im(\theta)/2}$. So if we write $z_n = r_n e^{i\theta_n}$, then

$$ r_n = r^{2 - 2^{-n}}, \qquad \theta_0 = \theta, \qquad \theta_{n+1} = \theta + \frac{1}{2}m(\theta_n) $$

As a consequence,

  • If $|\theta| \leq \frac{\pi}{2}$, then we can inductively show that $\theta_n = (2 - 2^{-n})\theta \in (-\pi, \pi)$ and hence

    $$ z_n \xrightarrow[n\to\infty]{} r^2 e^{2i\theta} = z^2. $$

  • Now consider the case $\theta = \frac{2\pi}{3}$. Then we can show that $(\theta_n)$ has 3 limit points $\frac{4 \pi}{21}, \frac{16\pi}{21}, \frac{21 \pi}{21}$. This in particular tells that $z_n$ does not converge as $n\to\infty$. This kind of behavior is general for $\theta \in (\frac{\pi}{2}, \pi]$, as we see from the graph of $\theta$ versus limit points of $(\theta_n)$.

    $\hspace{9em}$ Graph of limit points of $(\theta_n)$

    This tells that $i\sqrt{i\sqrt{i\sqrt{i\cdots}}} = i^2 = -1$ is sort of an 'edge case'.


By the same way it means: $$i^{1+\frac{1}{2}+\frac{1}{4}+\cdots}=i^2=-1.$$

  • $\begingroup$ Can someone to explain me what is a different between my solution and the Donald Splutterwit's solution? By the way, I posted my solution before. Thank you! $\endgroup$ Mar 2, 2018 at 13:12
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    $\begingroup$ Maybe it's more visually arresting? $\endgroup$
    – LarsH
    Mar 2, 2018 at 13:41
  • $\begingroup$ You have my sympathies. :-) But since you asked: not only is the other answer more visually arresting, this one also requires the reader to scroll up and re-read the question. (E.g. it starts with “by the same way”: which way?) (And even the meaning of “it” refers to an expression that occurs in the question.) The other answer makes a statement that stands independently (you can read it and gain something from it even if you have forgotten the question). Such answers tend to be easier to read, especially if the reader has read other answers between reading the question and this one. $\endgroup$ Mar 2, 2018 at 22:45

One way to approach this fixed-point problem rigorously is to use the polar form of complex numbers. Consider the action of the mapping $$z\mapsto a\mathrm{e}^{\mathrm{i}\alpha}\sqrt{z}$$ when $z=r\mathrm{e}^{\mathrm{i}\phi}$ is expressed in polar form, $r>0$, $a>0$, $-\pi/2\leq\alpha\leq\pi/2$, $-\pi<\phi<\pi$. Under this mapping $$\begin{align}\ln r&\mapsto \tfrac{1}{2}\ln r+\ln a\\ \phi&\mapsto \tfrac{1}{2}\phi+\alpha\end{align}$$ Since this is a contractive mapping, it has a unique fixed point which must be $(\ln r,\phi)=(2\ln\alpha,2a)$. The result follows from letting $a=1$ and $\alpha=\tfrac{\pi}{2}$.


The two highest-voted answers (as of writing) are incorrect because they do not show that the expression even has a limit, only that if the limit exists then it is −1. – Rahul

We have the sequence

$$ a_0 = i,\quad a_{n+1} = i \sqrt{a_n}. $$

I think the other answers have sufficiently covered that the argument of each element of the sequence lies in $(0, \pi)$, so we can be sure that we're always taking the principal square root. They also have shown clearly enough (except for off-by-one errors) that

$$ a_n = \exp(i\pi - i\pi/2^n).$$

I assert that the limit of this sequence is $-1$. For any $\varepsilon>0$ there exists $N$ such that $|-1 - a_n| < \varepsilon$ when $n>N$.

$$| -1 - a_n |=| -1 - \exp(i\pi)\exp(-i\pi/2^m) |=| -1+\cos(\pi/2^n)+i\sin(\pi/2^n)|$$ which is less than or equal to

$$ |1 - \cos(\pi/2^n)| + |\sin(\pi/2^n)|. $$

$1-\cos(x)<x$ for all $x>0$, as is $\sin(x)<x$. So the above is less than $2\pi/2^n$. Then for $N > \log_2(2\pi/\varepsilon)$ we have $|a_n+1|<\varepsilon$.


Let $$x=i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }$$ $$\implies x^2=-1i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }$$ $\implies x^2=-x$ $\implies x^2+x=0$ $\implies x(x+1)=0\implies x=0\; \text{or} -1$ since $x$ cannot be $0$, hence $x=-1$


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