Let $C$ be the simple closed contour in a complex plane. If $P(z)$ is a polynomial with no root on the curve $C$, show that if $f(z)=\frac{P'(z)}{P(z)^2}$:
$$ \int_C \frac{P'(z)}{P(z)^2} dz = 0$$
$\underline{Attempt}$
The theorem I recall requires continuity of $f$ in the domain.
I was thinking I could use the continuity property of $f(z)$ on the domain and conclude the integral (on the curve) is $0$. I am not sure if this approach is correct considering that $f(z)$ might have discontinuities in the contour but not on the curve.
Let $Q(z)=-\frac 1 {P(z)}$. Then the given intergal is $\int_C Q'(z)dz$. By the definition of Contour integral the integral of any derivative over a closed contour is alwasy $0$. [This does not requite anayticity of $Q(z)$. The fact that $P(z) \neq 0$ in a neighborhood of $C$ is good enough].
[Suppose $C$ is given by $\gamma :[a,b] \to \mathbb C$. Then $\int_C Q'(z)dz=\int_a^{b} Q'(\gamma (t))\gamma'(t)dt=Q(\gamma (b))-Q(\gamma (a))=0$ since $\gamma (b)=\gamma (a)$].