I find hard to prove the following result : if $F=\frac PQ$ is a rational function, and if $a$ is a pole of $F$ of order $\alpha$, then there are reals $M>0$ and $\nu>0$ such that for all $t\in\mathbb{C}$ verifying $\left|t\right|>M$, equation $F(z)=t$ has exactly $\alpha$ zeros in the disc $B(a,\nu)=\left\{z\in\mathbb{C}, \left|z-a\right|<\nu\right\}$.
I tried this : let $Q=(X-a)^\alpha Q_1$, with $Q_1(a)\ne0$, the equation $F(z)=t$ can be written $$P(z)-t(z-a)^\alpha Q_1(z)=0 \iff (z-a)^\alpha = \frac1t \frac{P(z)}{Q_1(z)}$$ $\frac{P(a)}{Q_1(a)}\ne0$, so fraction $\frac{P}{Q_1}$ doesn't cancel in the vicinity of $a$. The equation looks "just like" equation $$(z-a)^\alpha = \frac1t \frac{P(a)}{Q_1(a)}$$ which clearly has exactly $\alpha$ solutions for $t$ sufficiently large (so $M$ has to be fixed AFTER $\nu$), but I don't know how to translate this to the actual equation.
A tip, anyone ?
Thanks.