I've been brushing up on my complex analysis recently, and I've come across a problem that's stumped me: What are the real and imaginary parts of $$\sqrt{i+\sqrt{i+\sqrt{i+\sqrt{i+\cdots}}}} ?$$

I really don't know how to start this problem; this is my first encounter with nested roots.

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    $\begingroup$ What is $x^2-x-i=0$ ? $\endgroup$
    – Someone
    Commented May 3, 2015 at 10:32
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    $\begingroup$ One problem I can see is : what is for you $\sqrt{z}$ with $z \in \mathbb{C}$? This need to be rigouriously defined before asking this question $\endgroup$
    – Tryss
    Commented May 3, 2015 at 10:41
  • $\begingroup$ Mann squaring is not an equivalence operation. $\endgroup$ Commented May 3, 2015 at 12:22

3 Answers 3


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.

  • $\begingroup$ I just did quick calculation , though might be wrong. But the root seems $x=\frac{1}{2}\pm \frac{1}{4} [(\sqrt{17}+1)^{1/2}+i (\sqrt{17}-1)^{1/2}]$ Probable the negative sign, doesn't seem the answer. $\endgroup$
    – Someone
    Commented May 3, 2015 at 10:40
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    $\begingroup$ I don't see any problem. A quick numerical check suggests that the sequence approaches the root of the given quadratic in the upper half-plane. $\endgroup$ Commented May 3, 2015 at 10:54
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    $\begingroup$ I upvoted this because it was the first answer showing awareness of the few intricacies involved in this question. I can only speculate. May be the downvoter wants to see a proof for the existence of the limit or something? IDK. $\endgroup$ Commented May 3, 2015 at 11:32
  • $\begingroup$ Squaring is not a equivalence. You may as well be getting $x^2-i=-x$. $\endgroup$ Commented May 3, 2015 at 12:18
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    $\begingroup$ @mathreadler It does not need to be an equivalence for this solution as written (or sketched anyway); squaring does introduce a "false solution", but addressing this is anyway exactly the content of (3). $\endgroup$ Commented May 3, 2015 at 12:26

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.

  • $\begingroup$ I hope it's clear to all that there's not much magic in the choice of $3/5$ as a bound for the norm in the definition of $D$. It is just a nice round number a bit larger than $1/2$. The same applies to the upper bound $2$. $\endgroup$ Commented May 3, 2015 at 11:30
  • $\begingroup$ Beautiful solution! $\endgroup$ Commented May 3, 2015 at 12:35
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    $\begingroup$ I'm a bit unhappy with my end game actually. While typing it I thought Bolzano-Weierstrass, continuity, and uniqueness of the fixed point would get the job done. But I don't think those alone suffice. For example some iterates of $f$ could have other fixed points leaving open the possibility of different subsequences converging to different limits. So I need the contraction property. But then the BW-business became kinda moot. I don't wanna bump this any more, so, having aired these misgivings, I will just let this sit. $\endgroup$ Commented May 4, 2015 at 7:11
  • $\begingroup$ Interestingly enough, the magnitude of the derivative around the second solution is less than one, but it is an unstable fixed point, from numerical experiments. Any idea on how to prove this? $\endgroup$ Commented Apr 27, 2019 at 13:38

We can write $$z^2 = i+z$$

Let $z = a+bi$

Now we have $$(a+bi)^2 = i+(a+bi)$$

$$(a^2-b^2)+2abi = a+(b+1)i$$

We have two equations $$a^2-b^2 = a$$ And $$2ab = b+1$$

Use wolfram to do algebra, appears that there are two solutions



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