In complex analysis notes I saw the following argument:
Let $a,b$ be holomorphic s.t. $b(z_0)=0$, $b'(z_0)\neq 0$ and $a(z_0)\neq 0$, for some $z_0$, then:
$Res(\frac{a(z)}{b(z)}, z_0)=\frac{a(z_0)}{b'(z_0)}$.
I don't see where the assumption $a(z_0)\neq 0$ is used. Here how I proved it:
Since $z_0$ is a simple root for $b$, there exists an analytic function $g$ s.t. $b(z)=(z-z_0)g(z)$ and $g(z_0)\neq 0$.
Now, $g(z_0)=lim_{z\to z_0}\frac{b(z)-b(z_0)}{z-z_0}=b'(z_0)$. Therefore,
$Res(\frac{a(z)}{b(z)}, z_0)=\int_{\gamma}\frac{a(z)}{b(z)}dz=\int_{\gamma}\frac{\frac{a(z)}{g(z)}}{(z-z_0)}dz=\frac{a(z_0)}{g(z_0)}=\frac{a(z_0)}{b'(z_0)}$, where $\gamma$ is some circle around $z_0$ and the third equality is from Cauchy's Integral Formula.
What do I do wrong?
Thnks!
If $a(z_0)\neq 0$ then $z_0$ may not be a pole of $\dfrac{a(z)}{b(z)}$ always.
Consider for example the function $\dfrac{z^2}{z}$.