Contour integral

Question

Suppose $f(z) = f(0) \prod_{\rho}^{} (1 - \frac{z}{\rho})$ where $\rho$ is a root of $f(z)$. Evaluate the contour integral $\int_{\gamma}^{} \frac{f'(z)}{f(z)} dz $ where $\gamma$ is the rectangle $\gamma = (a , a + b, a + b + it, a+ it)$ traversed anti-clockwise and the roots of $f(z)$ are contained inside the rectangle.

Here is what I have so far:

$\frac{f'(z)}{f(z)} = - \sum_{\rho} \frac{1}{\rho}\frac{1}{1-\frac{z}{\rho}} = \sum_{\rho} \frac{1}{z - \rho} $

Hence

$\int_{\gamma}^{} \frac{f'(z)}{f(z)} dz = \int_{\gamma}^{} \sum_{\rho} \frac{1}{z - \rho} dz = \sum_{\rho} \int_{\gamma}^{} \frac{1}{z - \rho} dz $

Now my question is, can I apply Cauchy's integral formula at this stage, or do I need to parametrize the contour integral?

Answer 1

By Cauchy's integral formula, $\int_{\gamma} \frac{1}{z - \rho} dz=2\pi i\cdot 1.$ Therefore $$\int_{\gamma} \frac{f'(z)}{f(z)} dz= \sum_{\rho} \int_{\gamma}^{} \frac{1}{z - \rho} dz=2\pi i \cdot n$$ where $n$ is the algebraic number of roots inside the simple closed curve $\gamma$ traversed anti-clockwise.

Answer 2

Hint:

$$f(z)=f(0)\prod_{\rho}\left(1-\frac z\rho\right)\implies f'(z)=f(0)\sum_{\rho}\prod_{\tau\neq\rho}\left(-\frac1\rho\right)\left(1-\frac z\tau\right)\implies$$

$$\frac{f'(z)}{f(z)}=-\sum_{\rho}\frac1{\rho\left(1-\frac z\rho\right)}=\sum_{\rho}\frac1{z-\rho}$$

If you knew the Argument Principle then this exercise is way easier...