For $n$ odd ( e.g. with $n\equiv 1\mod 4 )$ I seek a solution $f(n)$ for this simple equation in the complex numbers $$(-1)^{f(n)}2^{\frac{n-1}{2}}=-\frac i2(1+i)^{n+1}$$ $f(n)$ is probably an integer valued funcion.
As I suppose it will be helpful to change into exponential representation and then take logarithm - but I am not firm about multivaluedness.
On
If $f(n)$ is indeed integer valued, then $(-1)^{f(n)} \in \{-1,1\}$. If $f(n)$ is a solution, then all $f(n) + 2k\ \ \forall k \in \mathbb{Z}$ are solutions, as well. The fact that the equation is of complex nature, doesn't change that. So for every n, you basically have only two possible solutions: $f(n)$ even or odd.
We have $$(-1)^{f(n)} = -i \frac{(1+i)^{n+1}}{\sqrt{2}^{n+1}}$$
Note that:
$$(1+i) = \sqrt{2}·e^{\large i\pi/4}$$
Therefore:
$$(-1)^{f(n)} = -i·e^{\large(n+1)i\pi/4}$$
Since $-i = e^{\large -i\pi/2}$ and $-1 = e^{\large i\pi}$
Then we want:
$$e^{\large i\pi ·f(n)} = e^{\large(n-1)i\pi/4}$$
So choosing $f(n) = \frac{n-1}{4}$ would work.
Note that since $n \equiv 1 \pmod 4$ then $f(n)$ is indeed an integer valued function.