# How many solutions does $z^{\sqrt{2}}=2$ have?

Let's explore: $$f(z)=z^{\sqrt2}-2$$

Question

How many solutions does $$f$$ have in the complex plane? Finite, Countable or Uncountable?

Exposition One may define $$e^{a+bi}=e^{a}(\cos(a)+i\sin(b)), \hspace{2 cm }c^{a+bi}=e^{\ln(c)(a+bi)}$$ Next we may use the continuity of our function and the rationals to extend our function from the rationals to the reals to complex numbers. I believe this is the way Walter Rudin does this. I am happy to read answers to this question which define $$e^z$$ in other ways. There are multiple ways to do this but one hopes for some fidelity of the conclusions.

On the reals, this should have the solution $$2^{1/\sqrt{2}}$$. But then what of the complex solutions?

$$(re^{\theta i})^\sqrt{2}=2 \implies r=2^{1/\sqrt{2}}$$ and $$(e^{\theta i})^{\sqrt{2}}=e^{2\pi n i}$$ for every $$n\in \mathbb{Z} \implies \theta=\frac{2 \pi n}{\sqrt{2}}$$ And now I suspect we have found all of the countably many solutions: $$z= 2^{\frac{1}{\sqrt{2}}}e^{2n\pi i/\sqrt2}$$ I seem to have countably solutions which are a dense subset of the circle $$|z|=2^{1/\sqrt{2}}$$. But these countable solutions seem to point to the answer that there should be uncountably many solutions. Our multivalued function $$f(z)$$ which should be interpreted as continuous (in each branch cut) should then take on zero along the entire circle by continuity. This is certainly not the case in the image produced below where we can see the zero of the function at $$2^{1/\sqrt2}+0i$$ where the different colors converge. I assume this is because we are looking at different branch cut.

Motivations: I was exploring Gelfond-Schneider and noting that these solutions could be written $$z=\left(-\sqrt{2}\right)^{\frac{2}{\sqrt{2}}}$$ I wanted to argue that all the solutions could be argued transcendental. But then I had a worry that there should be more solutions than I expected. I figured that this case might lead to some insight about whether we can make any claims about the solution sets of more exotic looking things like $$z^\sqrt{2}+z^{\sqrt{3}}=7$$

One can produce an image of our function:

• $f$ is not well-defined over $\Bbb C$, because the logarithm function ($\ln(c)$ in your definition) is multi-valued over $\Bbb C$. Commented Sep 3 at 19:49
• I get that $f$ is multivalued: My point is isn't one of those values zero? That is can't we say $0\in f(z)$? But how do we get from multivalued to not well-defined? Aren't the sets $f(x)=f(y)$ whenever $x=y$? Commented Sep 3 at 20:01
• I lost a good answer! The deleted answer explained that these "zeros" weren't next to each other on the appropriate Riemann surface only in the projection onto the plane. So there are countably many "solutions" on the Riemann surface. Also it had a link: samuelj.li/complex-function-plotter/#z%5E(sqrt(5))-2 Commented Sep 3 at 20:42
• countable solutions - there is a very general theorem that states any multivalued analytic function in the plane (eg $z^{\sqrt 2}$) has at most countable germs at any point so in particular if there would be uncountable roots there would be a finite accumulation regular point of them etc - in general a multivalued function is the collection of all analytic continuations of a given power series - here one can just use the power series of $e^{\sqrt 2 \log z}$ near $1$ with the principal branch of the logarithm to generate all possible branches of $z^{\sqrt 2}$ at all regular points ) Commented Sep 3 at 22:42
• here of course you can actually write down the solutions in terms of two integral parameters Commented Sep 3 at 22:44

The zeros of $$f$$ differ in argument by $$\theta=\sqrt{2} \pi>\pi$$ and since you chose the principal branch (of $$\log z$$) in your plot the only zero on it is the zero at zero and the next isn't on it since you only plot $$f$$ for $$z$$ with argument $$(-\pi,\pi)$$ and $$\pm\theta\not\in(-\pi,\pi)$$. You can for example choose $$\sqrt{5}$$ instead of $$\sqrt{2}$$ so in this case you can see $$3$$ zeros of $$g(z)=z^{\sqrt5}-2$$ in the principal branch plot since $$2/\sqrt{5}<1$$.

• So how many solutions of $f$ are there in $\mathbb{C}$? Commented Sep 3 at 20:00
• @Mason It is not a holomorphic function on $\mathbb{C}$. You could define it for $\mathbb{C}$ but then it wouldn't be holomorphic or even continuous and you would need to precisely specify which branches of the logarithm you are choosing for $z\in\mathbb{C}$.
– user1385442
Commented Sep 3 at 20:01
• It's not a function into $\mathbb{C}$ but it is a multivalued function. So for each $z\in \mathbb{Z}$ we have a set of values from $\mathbb{C}$. Maybe my question is something like this for how many $z\in \mathbb{C}$ do we have that $0\in f(z)$? That is, zero is a member of this set. Thanks for your answer and the patience. Commented Sep 3 at 20:05
• We can't just define the thing in terms of sets? I have a suspicion that in one of these branches we have have that $f(z)=0$ when we take $z= 2^{\frac{1}{\sqrt{2}}}e^{2n\pi i/\sqrt2}$. But i suppose if these appear in different branches (for different $n$) we cannot use continuity and we're back to countably many solutions; where solutions just means $z$ such that $0$ appears in the set $f(z)$ Commented Sep 3 at 20:13
• @Mason Yes, nice!
– user1385442
Commented Sep 3 at 20:22

I read a pretty decent answer which was deleted which I suppose is a good enough excuse to write down an answer.

If we consider this $$f$$ on the domain of the complex numbers we get a multivalued function but the more appropriate tool would be a Riemann surface because it's more equipped to differentiate between equivalent theta. On this surface we have a genuine function which has countably many zeros which are not actually next to each other so the continuity argument fails.

On the principal branch of our function $$f$$ we only have one zero and each zero has another zero with an angle of $$\frac{2\pi}{\sqrt{2}}$$ between them and because this value is greater than $$\pi$$ we don't see the next zero but had we considered $$g(z)=z^{\sqrt5}-2$$ we can spot two more zeros (with $$\theta= \pm \frac{2\pi}{\sqrt{5}}$$) in the graph of principal branch.

It is true that the zeros of $$f$$ project into a dense subset of the circle $$z=2^{1/\sqrt{2}}$$ in the plane. And less exotic looking things like polynomials such as $$z^{13}-2$$ will have countably many solutions on the Riemann surface but only 13 solutions when projected into the plane.

What remains is to consider what happens with the more exotic forms like $$h(z)=z^\sqrt{2}+z^\sqrt{7}-5$$. In the comments Conrad writes that this can be argued to have at most countably many solutions but I'm not sure how we go about showing that it has infinitely many solutions. It might be nice to see what the solutions to this look like when projected into the plane.