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I was working on the following nested square root problem:

Let $a \in \mathbb R ^+$, what is the value of: $$\sqrt{a \sqrt{a \sqrt{a}}}...$$

I concluded that the answer is $a$ and then I thought about generalizing this to the complex numbers. The question became:

Let $z \in \mathbb C$, what is the value of: $$\sqrt{z \sqrt{z \sqrt{z}}}...$$

I did the following:

Let $x \in \mathbb C$ with $x =\sqrt{z \sqrt{z \sqrt{z}}}... $. Then: $$\sqrt{z \sqrt{z \sqrt{z}}}... = \sqrt{z} \sqrt{\underbrace{\sqrt{z \sqrt{z \sqrt{z}}}...}_x}$$

So we concude that $x = \sqrt{z} \sqrt{x}$ or: $$x^2 = |z| |x| \ \ (1)$$.

Let $x = r e^{i \phi}$ and $ z = p e^{i \alpha}$. Then, eq. (1) becomes:

$$r^2e^{i 2 \phi} = pr$$

$$r = 0\vee r = p e^{- i 2 \phi}$$

We know that $r \neq 0$ so we have that $r = p e^{- i 2 \phi}$. because $x = r e^{i \phi}$ if we substitute this we get:

$$x = p e^{- i \phi}$$

So this concludes that:

$$\left| \sqrt{z \sqrt{z \sqrt{z}}}... \right| = |z|$$

My question is: I was able to find the absolute value of the solution, now to fully have a solution for this problem I need the argument of it. How can find the argument of the solution?

656475
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  • https://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation/589293#589293 – lab bhattacharjee Feb 13 '21 at 18:34
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    That isn't helpful @labbhattacharjee, I'm working with complex numbers, and As I said, I already did the case where $z$ is a positive real number (the topic of that link) – 656475 Feb 13 '21 at 18:38
  • Do you know how to define the square root function for complex numbers? – anon Feb 13 '21 at 18:46
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    $\sqrt{z \sqrt{z \sqrt{z}}}...$ is plainly the noise done by a sleeping square root... – Jean Marie Feb 13 '21 at 18:48
  • Yes @runway44. The square root of a number $z$, is a number denoted by $\sqrt{z}$ such that that number squared is $z$. – 656475 Feb 13 '21 at 18:50
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    That sentence technically doesn't make sense in English, for a pedantic reason: when you say "the" square root, you're implying there is only one value, but of course there are two numbers that square to any nonzero $z$. For positive values of $z$, we pick the positive root to be "the" (principal) square root. Do you know what we do for complex numbers? For instance, what would you say $\sqrt{i}$ and $\sqrt{-i}$ are? Can you express them as $a+bi$ explicitly? – anon Feb 13 '21 at 18:53

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Over the entire complex field, it's not possible to consistently (continuously) define $\sqrt .$ everywhere. It's the same problem as defining a logarithm : you have to cut somewhere.

So your question won't have a definite answer. My guess is : for every phase $\varphi$, there is a definition of $\sqrt .$ such that your answer is $|z|e^{i\varphi}$.

The idea is that the series $\sum \frac {x_i} 2$ for $(x_i) \in \{0,1\}^{\mathbb N}$ can converge to anything in $[0,1]$ so by defining the cut to realise the right series $(x_i)$ we can add any phase we want.

Hugo Manet
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  • Picking a specific sequence $(x_i)$ amounts to picking a potentially different branch cut for each individual application of the square root. – anon Feb 13 '21 at 18:57