How one can find the positive integer $k$ in term of complex logarithm of $f(s)/f(3-s)$ if $f(3-s)$ is not zero. Let us consider the non constant complex analytic function $f$ on all $ℂ$ verifying the functional equation:
$$f(s)=(-1)^{k}f(3-s)$$
for all complex number $s$ where $k$ is a fixed positive integer.
My question is: How one can find the positive integer $k$ in term of complex logarithm (What is the complex logarithm of $(-1)^{k}$) of $f(s)/f(3-s)$ if $f(3-s)$ is not zero.

The positive integer $k$ is for the most part unrecoverable from the functional equation, because most of the information is gone after raising $-1$ to the power of $k$. Using the complex logarithm won't help; all you can tell is whether $k$ is even or odd. $k$ is even if
$$f(s)=f(3-s)$$
for some $s$ such that $f(s)\neq 0$, and odd if
$$f(s)=-f(3-s)$$
for some $s$ such that $f(s)\neq 0$.
The logarithm is multivalued, and it will yield all of the odd or even integers if you try to solve for $k$, depending on whether $k$ is even or odd.
In more detail, if you try to solve for $k$, you would write this as
$$e^{ki\pi} = \frac{f(s)}{f(3-s)}$$
so
$$k i\pi = 2m\pi i+\log\frac{f(s)}{f(3-s)}$$
as $m$ ranges over all integers. (If you know one logarithm, any other logarithm differs by an integer multiple of $2\pi i$).
Thus
$$k = 2m+\frac 1{i\pi}\log\frac{f(s)}{f(3-s)}$$
But here we know that if $k$ is even, then $(-1)^k=1$, so the principal logarithm of $\frac{f(s)}{f(3-s)}=0$, so $k=2m$ for some $m$. If $k$ is odd, then 
$$\frac{f(s)}{f(3-s)}=-1$$
so the principal logarithm is $i\pi$. Dividing by $i\pi$, we get that
$$k=2m+1$$
so $k$ can be any odd integer.
