Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$? 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?
 A: \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*}
A: 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 $x=-1$
A: 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}
A: 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 $(\theta_n)$.
$\hspace{9em}$ 
This tells that $i\sqrt{i\sqrt{i\sqrt{i\cdots}}} = i^2 = -1$ is sort of an 'edge case'.
A: By the same way it means:
$$i^{1+\frac{1}{2}+\frac{1}{4}+\cdots}=i^2=-1.$$
A: 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 $\alpha=\tfrac{\pi}{2}$.
A: 
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 $|a_n+1|<\varepsilon$.
