Remember that entire taylor series are coo everywhere. ( infinitely differentiable for all finite complex )
Let $f(z,1)$ be an entire periodic function with $f(0,1)=f(1,1)=1$ and period 1.
And $f(z,1)$ is not identically 1 for all $z$.
We will prove that for complex b with $arg(b) <> 0$ , the only solution to the equations is $f(z,1) * b^z$ and hence the proof follows.
Let k and n be positive integers.
$f(0) = 1$
$f(z+k) = b^k f(z)$
$f$ = entire
Take the derivative of the equation $f(z+k) = b^k f(z)$ on both sides
$f ' (z+k) = b^k f ' (z)$
$ f '' (z+k) = b^k f '' (z)$
And in general
f^(n) (z+k) = b^k f^(n) (z)
Hence because of taylors theorem we must conclude
$f(z) = f(0) * f(z,1) * b^z$ in the neighbourhood of 0.
But since f is entire it must be true everywhere and $f(0) = 1$ hence
$f(z) = f(z,1) b^z$
for all z.
If $arg(b) <> 0$ then the period of $b^z$ does not have $Re <> 0$ and hence $b^z$ is unbounded on the strip.
If $f(z)$ needs to be bounded and $b^z$ is not bounded , this implies that $f(z,1)$ needs to be bounded.
But this is impossible since $f(z,1)$ has a real period and is entire , it must be unbounded on the strip.
( remember $f(z,1) =/= 1$ everywhere by definition )
The product of two functions unbounded in the same region must be unbounded in that region.