Is $|f(z)|=|f(z^*)|$ true, where $z$ is a complex number and $z^*$ is it's conjugate?

Question

I recently stumbled upon the relation, $$|i!|=\sqrt{\frac{\pi}{\sinh\pi}}$$ And while trying to prove it, I realized that the following is true, $$|\Gamma(1+i)|=|\Gamma(1-i)|$$ I was just wondering if this could be generalized for any complex function and if there is any proof for that.

Answer 1

Definitely not true in general. Counterexample: let $f(z)=\operatorname{Arg}(z)+\pi/4.$ Then $|f(1+i)|=\pi/4+\pi/4=\pi/2,$ but $|f(1-i)|=-\pi/4+\pi/4=0.$

Answer 2

wrong:
consider $f(z)=Im(z)+1$,
yielding $f(i)=2$, $f(i*)=f(-i)=0$.

--- rk

Answer 3

Among analytic function the claim is certainly true if $f$ maps real numbers to real numbers, like $f(z)=2z+1$. But there are also analytic functions where real numbers are not mapped to each other, like $f(z)=\exp(iz)$, for which the claim is wildly off base. With the latter function $|f(-i)|$ is about seven times as large as $|f(i)|$.

Not only is the claim disproved, we sometimes put the disproof to use. Certain integrals form $-\infty$ to $+\infty$ can be evaluated using the Residue Theorem by incorporating the integration path into a closed contour. When we close the contour in one way we go bad because the closure dominates the integral we wanted, but if we close the contour the other way it works because the condition $|f(z)|\ne |f(\overline{z})|$ makes the alternate closing path integral small like we want it to be.

Answer 4

Not in general. Consider $f(z)=z+i$