Prove that $\oint _{|z|=R} (f-g)' dz = 0$ (Residue Theorem)

Question

I know that $f$ and $g$ have a pole or order $k$ in $z=0$. $f-g$ is holomorph in $\infty$.

I need to prove that:

$$\oint_{|z|=R} (f-g)' dz = 0$$

Any help?

Note: $f$ and $g$ only have a singularity in $z=0$

Answer 1

If $h$ is holomorphic, then $h'$ exists and is holomorphic. If $h$ is holomorphic, then $\oint_C h dz = 0$ for all circles $C$. Take $h= f-g$, $(f-g)'$ respectively.

Answer 2

As Git Gud alluded to in their comment, simply parameterise the curve and calculate the integral directly. The usual parameterisation is $\gamma:[0,2\pi]\rightarrow\mathbb C:\ t\mapsto Re^{it}$. Letting $h=f-g$, we have $$\oint_Ch'(z)\ \mathrm dz=\int_0^{2\pi}h'(\gamma(t))\gamma'(t)\ \mathrm dt=\int_0^{2\pi}(h\circ\gamma)'(t) \mathrm dt\\=(h\circ\gamma)(2\pi)-(h\circ\gamma)(0)=h(R)-h(R)=0.$$