Ok, I give it another try:

Proposition.

Let , , and being a domain (open and connected) of definition and being a domain (open and connected) of values for a holomorphic function.

Let be a holomorphic function on , , such that

(0) .

(1)

(2)

(U) has bounded real part.

Then for every other on holomorphic satisfying (0), (1), (2), (U).

Proof.

has bounded real part on . We consider and and so to be holomorphic on the same Riemann surface . is a 1-periodic function, holomorphic on . As it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip otherwise it is a constant. Now has bounded real part and hence can not take on every value, so and .

Proposition.

Let , , and being a domain (open and connected) of definition and being a domain (open and connected) of values for a holomorphic function.

Let be a holomorphic function on , , such that

(0) .

(1)

(2)

(U) has bounded real part.

Then for every other on holomorphic satisfying (0), (1), (2), (U).

Proof.

has bounded real part on . We consider and and so to be holomorphic on the same Riemann surface . is a 1-periodic function, holomorphic on . As it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip otherwise it is a constant. Now has bounded real part and hence can not take on every value, so and .