Prove that for a holomorphic function $f$ on $\mathbb{C}$, if $-f(z) = f(\frac{1}{z})$, then the residue of $f$ at 0 is 0.
Note: $f$ isn't defined on 0. I'm having a bit of trouble getting this problem because I can't use what I normally use (Cauchy's integral formula) to calculate the residue. I also thought about calculating the Laurent series for $f$, dividing by $z$ and then looking at the $n = -1$ coefficient, but that is basically the same thing.
I know I can look at the function $g = f(\frac{1}{z}) + f(z), g = 0 $, but I don't think it leads anywhere. Also, I have a hunch that $f$ can only be the 0 function. Is this correct? Any help would be appreciated!
$g(z) = f(z) + f(\frac 1z) = 0$. $g(z)$ is holomorphic also, so $g'(z) = -\frac 1{z^2} f'(\frac 1z) + f'(z) = 0$ for all $z \neq 0$.
This implies that $z^2f'(z) = f'(\frac 1z)$
Now write $f(z) = \sum c_i z^i$, it's laurent expression in a neighborhood of $0$
What is the known term on the left hand side? Is $-c_{-1}$
And on the right hand side? is $c_{-1}$
Hence $-c_{-1} = c_{-1} \implies c_{-1} = 0$
Note the the Laurent series is differentiable term by term inside it's disk of convergence. So from $f(z) = \sum c_i z^i$ you get $f'(z) = \sum ic_i z^{i-1}$, $f(1/z) = \sum c_i z^{-i}$ , $f'(1/z) = \sum -ic_iz^{-i-1}$