complex analysis: If $f$ is analytic and $\operatorname{Re}f(z) = \operatorname{Re}f(z+1)$ then $Im\;f(z) - Im\;f(z+1)$ is a constant I am having trouble deciphering the reason behind a line in a complex analysis textbook (Complex made Simple by Ullrich, page 360 5 lines down in Proof of Theorem B, for those who are interested).  
Basically it says that since $f(z)$ is a holomorphic function with  $\operatorname{Re} f(z+1) = \operatorname{Re}f(z)$, that for all $z$:
$$\begin{align}
f(z+1) - f(z) &= \operatorname{Re}f(z+1) + i\cdot\operatorname{Im}f(z+1) - \operatorname{Re}f(z) - i\cdot\operatorname{Im}f(z)\\
&= i\cdot\operatorname{Im}f(z+1) - i\cdot\operatorname{Im}f(z)\\
&= i\cdot \text{constant}.
\end{align}$$
I do not understand why we must have that for all $z$ $\quad\operatorname{Im}f(z+1) - \operatorname{Im}f(z) = $ a constant.
Can anyone help?  
Thanks
 A: I will give it a try, feel free to correct me. Since $f(z)$ is analytic then $f(z+1)$ is analytic aswell. write $f(x)= u + iv$ and $f(z+1)=u^{\ast} + iv^{\ast}$ so for $f(z)$ we have by the Cauchy-riemann equations: $$\frac {du}{dx}=\frac {dv}{dy}$$ and for $f(z+1)$ we have: $$\frac {du^{\ast}}{dx} = \frac {dv^{\ast}}{dy}$$ Now since we are given the information $\Re(f(z)) = \Re(f(z+1))$ it follows that: $$\frac {du}{dx} = \frac {du^{\ast}}{dx} \implies \frac {dv}{dy} = \frac {dv^{\ast}}{dy}$$ Integrate the last expression above and you will obtain $v-v^{\ast} = C$ for some constant $C \in \mathbb{C}$, in other words, $\Im(f(z))-\Im(f(z+1))$ = C$
A: Here's a slightly different approach to the same idea as George Mouselli's proof. The function $z\mapsto f(z)-f(z+1)$ is analytic because $f$ is, and its real part is constant. The Cauchy-Riemann conditions then tell you that its imaginary part has zero partial derivatives in both directions, so it's constant too.
Alternatively, you could use the fact that a non-constant analytic function is an open map. But all values of $f(z)-f(z+1)$ are, by hypothesis, on the imaginary axis.  So this map can't be open and must therefore be constant.
A: That is wrong. Consider $f(z) = \cos 2 \pi z$, there $f(z + 1) = f(z)$, but $\Im(f(z))$ isn't constant. 
