I struggle with the following problem:
For a function $$f: \mathbb{C} \rightarrow \mathbb{R}~,$$ $f$ attains its maximum for $z_0= e^{i\pi/3}$, $f(z_0)=F_{max}.$
Assume we may use Taylor's theorem (identifying $\mathbb{C} =\mathbb{R}^2$) and write $$ f(z)= F_{max} + \underbrace{f^\prime(z_0)(z-z_0)}_{=0} + \frac{1}{2}f^{\prime \prime}(z_0)(z-z_0,z-z_0) + O(|z-z_0|^3).$$
Now I hope to get an error term of the form $$ F_{max} - f(z) = c(|z-z_0|^2) + O(|z-z_0|^3)~,~ c=const.~~~~~~~~~(1)$$
That would be true if the quadratic form $f^{\prime \prime}(z_0)$ would be just a multiple of the identity, which would in turn imply that $f$ takes the same values on small circles around $z_0$ (right?).
What I know is that $f(z)=f(\tilde{z})$ whenever the triangles $(0,1,z)$ and $(0,1,\tilde{z})$ are similar, for example $f(i)=f(1/2+i1/2)= f(1+i).$ But those values don't happen to lie on a circle around $z_0$, I think.
Can I use this information show that $f^{\prime \prime}(z_0)$ is just a multiple of the identity?
Or is there maybe another way to obtain equation $(1)$?
Thank you for helping.