Simple zero $a$ is equal to $\frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz$ for $f$ holomorphic within $C$?

Question

I'm struggling with this question:\

Show that the root $a$ which is a simple zero of $f(z)$ is given by $$a = \frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz, $$ where $f$ is holomorphic function in a simple closed contour $C$ and $f(z) \neq 0$ and $a$ is in $C.$ Also, $f(z) \neq 0$ for $z \in \{\mathbb{C}-a \} \ \text{in} \ C.$

I just know that since $a$ is a simple pole, then $f(a) = 0$ and $f'(a) \neq 0.$ Maybe I will use Residue theorem... but I don't know where to start.

Answer 1

Let $a \neq 0$. $\frac {zf'(a)} {f(z)}$ has a simple pole at $a$ and the residue is $\lim_{z \to a} \frac {z(z-a)f'(a)} {f(z)}=\lim_{z \to a} \frac {z(z-a)f'(a)} {f(z)-f(a)}$. By definition of the derivative $f'(a)$ this limit is simply $a$. [Divide numerator and denominator by $z-a$ and note that $\frac {f(z)-f(a)} {z-a} \to f'(a)$ as $z \to a$]. Hence, Residue Theorem gives the value of the integral divided by $2\pi i$ as $a$.

As pointed out in the comment below the result follows by Cauchy's Theorem when $a=0$ since there is no pole in that case.

Answer 2

By the residue theorem,$$\frac1{2\pi i}\int_C\frac{zf'(z)}{f(z)}\,\mathrm dz=\operatorname{res}_{z=a}\left(\frac{zf'(z)}{f(z)}\right).$$Since $a$ is either a removable singularity or a simple pole of $f$,\begin{align}\operatorname{res}_{z=a}\left(\frac{zf'(z)}{f(z)}\right)&=\lim_{z\to a}(z-a)\frac{zf'(z)}{f(z)}\\&=\lim_{z\to a}\frac{zf'(z)}{\frac{f(z)-f(a)}{z-a}}\\&=\frac{af'(a)}{f'(a)}\\&=a.\end{align}