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

That is wrong. Consider $f(z) = \cos 2 \pi z$, there $f(z + 1) = f(z)$, but $\Im(f(z))$ isn't constant. 
