Prove $\operatorname*{res}_{z=z_0} f(z)g'(z) = - \operatorname*{res}_{z=z_0} f'(z)g(z) $ 
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?
 A: 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}
A: 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 $0=\operatorname{res}\limits_{z=z_0} (f'(z)g(z) + f(z)g'(z)) = \operatorname{res}\limits_{z=z_0} f'(z)g(z) + \operatorname{res}\limits_{z=z_0} f(z)g'(z)$.
