Prove $\operatorname*{res}_{z=z_0} f(z)g'(z) = - \operatorname*{res}_{z=z_0} f'(z)g(z) $

Question

If $f$ has an isolated singularity at $z_0$ show that:

$$\operatorname*{res}_{z=z_0} f(z)g'(z) = - \operatorname*{res}_{z=z_0} f'(z)g(z)$$

Here is my proof using partial integration:

Proof

$$\begin{align} \operatorname*{res}_{z=z_0} f(z)g'(z) &= \frac{1}{2\pi i}\int\limits_{\partial B(z_0,r)} f(z)g'(z) \,\text{d}z \\ &\stackrel{\text{P.I.}}{=} \frac{1}{2\pi i} \left[ f(z)g(z)\right]^{z_1}_{z_0} - \frac{1}{2\pi i}\int\limits_{\partial B(z_0,r)} f'(z)g(z) \,\text{d}z\\ &= \color{red}{\frac{1}{2\pi i} \left[ f(z)g(z)\right]^{z_1}_{z_0}} - \operatorname*{res}_{z=z_0} f'(z)g(z)\\ &= \ldots \end{align}$$

Which proves the theorem if $\color{red}{ \left[ f(z)g(z)\right]^{z_1}_{z_0} = 0}$, but why would that be the case?

Answer 1

Because $z_0=z_1$, you shouldn't call the initial and final points of the curve like this since the center of the circle is $z_0$. To see it more clearly you may parametrize the curve. \begin{align} \operatorname*{res}_{z=z_0} f(z)g'(z) &= \frac{1}{2\pi i}\int\limits_{\partial B(z_0,r)} f(z)g'(z) \operatorname dz \\ &= \frac{1}{2\pi i}\int_{0}^{2\pi} f(z_0+re^{it})g'(z_0+re^{it})ire^{it} \operatorname{d}t \\ &= \frac{1}{2\pi i}\left(f(z_0+re^{it})g(z_0+re^{it})\left.\right\rvert_{0}^{2\pi} - \int_0^{2\pi}f'(z_0+re^{it})re^{it} g(z_0+re^{it})\operatorname dt \right)\\ &= 0 - \frac{1}{2\pi i}\int\limits_{\partial B(z_0,r)} f'(z)g(z) \operatorname dz\\ &= -\operatorname*{res}_{z=z_0} f'(z)g(z) \end{align}

Answer 2

Alternatively: Functions that have an anti-derivative on a small neighbourhood around $z_0$ have zero residue in $z_0$ and therefore $(f\cdot g)' = f'\cdot g + f\cdot g'$ implies $0=\operatorname{res}\limits_{z=z_0} (f'(z)g(z) + f(z)g'(z)) = \operatorname{res}\limits_{z=z_0} f'(z)g(z) + \operatorname{res}\limits_{z=z_0} f(z)g'(z)$.