Integral over the boundary of a rectangle This might be astupid question, but i really got stuck in it! So basically i have $F:\mathbb{C} \longrightarrow \mathbb{C}$ meromorphic, $\tau$ a complex number such that $Im(\tau)>0$and $0<Re(\tau)<1$, and  F has the periodicity $F(z+1)=F(z)=F(z+\tau)$. Suppose there are no zeros nor poles in the boundary of the rectangle whose vertices are $0,1,1+\tau,\tau$. I have to integrate $F'/F$ on the boundary of this retangle, and this should be zero. Parametrizing the segments in this way: $\gamma_1(t)=t$, $\gamma_2(t)=1+t\tau$,$\gamma_3(t)=1-t+\tau$,$\gamma_4(t)=(1-t)\tau$ (they all have domain $[0,1]$), and using the periodicity I obtain $$\int_0^1\frac{F'(t)}{F(t)}dt+\int_0^1\tau\frac{F'(t\tau)}{F(t\tau)}dt-\int_0^1\frac{F'(-t)}{F(-t)}dt-\int_0^1\tau\frac{F'(-t\tau)}{F(-t\tau)}dt$$. This should be zero because but i cannot explain to myself why he first and the third integral are opposite (and so the 2 and the 4).
 A: Why would you do such a messy parametrization of the rectangle (which is not a rectangle unless $\,\operatorname{Re}\tau=0\,$)? This looks as the beginning of elliptic functions and stuff, when you show how integrals of such functions as $\,F\,$ vanish on a fundamental paralleliped on the upper complex plane (and thus in fact $\,\operatorname{Im}\tau>0\,$) ..., but use periodicity of $\,F\,$ ! This function is the same on opposite sides of this parallelogram, which your integral "walks" on in opposite directions, so their values cancelate each other...as simple as that!
*Added*$\,\,\,$ Taking your parametrizations of the four segments, we get: $$\int_0^1\frac{F'(t)}{F(t)}dt+\tau\int_0^1\frac{F'(1+t\tau)}{F(1+t\tau)}dt-\int_0^1\frac{F'(1-t)}{F(1-t)}dt-\tau\int_0^1\frac{F'((1-t)\tau)}{F((1-t)\tau)}dt$$Put now $\,\,u=1-t\Longrightarrow du=-dt\,$ in the 3rd integral to get $$\int_1^0\frac{F'(u)}{F(u)}(-du)=\int_0^1\frac{F'(u)}{F(u)}du$$which is exactly your first integral, and now use that minus sign before the 3rd integral...do something similar for 2nd and 4th integrals.
