Does $\sqrt{i + \sqrt{i+ \sqrt{i + \sqrt{i + \cdots}}}}$ have a closed form?

Solution 1:

Hint If we denote $$x = \sqrt{i + \sqrt{i + \sqrt{i + \cdots}}},$$ then formally squaring and rearranging gives that $x$ ought to satisfy the quadratic equation $$x^2 - x - i = 0,$$ leaving two possibilities for $x$.

To solve the problem rigorously, you'll need:

1. to treat carefully the usual issue with nonintegral powers of complex numbers---this should include choosing a branch of the square root function---
2. to show that the sequence $\sqrt{i}, \sqrt{i + \sqrt{i}}, \sqrt{i + \sqrt{i + \sqrt{i + \cdots}}}$ converges (usually we interpret the nested root to be the limit of this sequence, and we can use this definition, together with an appeal to continuity. to make the above formal calculation rigorous), and
3. determine to which root the sequence in (2) converges (probably this depends on your choice in (1)).

Treating this problem in detail entails many subproblems of different types, from showing convergence to determining an expression for the real and imaginary components in terms of radicals. (Edit Jyrki Lahtonen has given a clear and detailed proof of the former in another answer to this question.) If you have more questions about a particular aspect, feel free to ask in the comments or just post a new question as appropriate.

Solution 2:

For the purposes of this exercise we can treat the complex square root, written in polar form like $re^{i\phi}\mapsto \sqrt re^{i\phi/2}$, as a continuous function from the closed upper half plane $\bar{H}$ to itself, so above we specify $r\ge0,\phi\in[0,\pi]$. If $z\in\bar{H}$ then so is $i+z$, and we get a function $f:\bar{H}\to\bar{H}$ that can be abbreviated as $$f(z)=\sqrt{i+z}.$$ Note that by specifying the phase of the square root as above I get a well defined function.

Let $$D=\{z\in\bar{H}\mid \frac35\le |z|\le 2\}.$$ If $z\in D$, then $1\le|z+i|\le3$ and consequently $\frac35<\sqrt{|z+i|}<2$, so we see that restricting $f$ gives us a continuous mapping $f:D\to D$. Furthermore, $f$ is holomorphic in the interior of $D$.

To make the question precise we can study the recursively defined sequence $x_1=i$, $x_{n+1}=f(x_n)$ for all $n\ge1$. The above considerations show that $(x_n)_{n\in\Bbb{N}}$ is a sequence of elements of $D$. Because the set $D$ is compact, it has a converging subsequence (Bolzano-Weierstrass). This is the complex analysis substitute for the real analysis drill of using the theorem of bounded monotonous sequences.

Using the other answers we see that the function $f$ has a unique fixed point in $D$. Differentiating the equation $$ f(z)^2=i+z $$ implies that for all $z\in D$ we have $$ 2f(z)f'(z)=1. $$ So $$ f'(z)=\frac1{2f(z)} $$ has absolute value $\le\dfrac56<1$ in all of $D$. Therefore $f$ is contractible in $D$, and that unique fixed point is the limit of the entire sequence.

Of course, by specifying different branches of the square root to be used at different steps we can get other sequences with other limits.