Prove of a residue being zero

Question

Im working on a question but I don't understand the final part:

"Let $f$ have an isolated singularity at $z_0$ ($f$ analytic in a punctured neighborhood of $z_0$). Show that the residue of the derivative $f'$ at $z_0$ is equal to zero."

The proof is as following:
Take the general form $f(z)= ...+a_{-2}(z-z_0)^{-2}+a_{-1}(z-z_0)^{-1}+a_0+a_{1}(z-z_0)^{1}+a_{2}(z-z_0)^{2}+ ...$
Then $f'(z)= ...-2a_{-2}(z-z_0)^{-3}-a_{-1}(z-z_0)^{-2}+a_{1}+2a_{2}(z-z_0)+ ...$
The latter one doesnt have a $(z-z_0)^{-1}$ term therefore the residue is zero.

I don't get the final part, why does not having a $(z-z_0)^{-1}$ term imply the residue is zero?

For computing a residue I use the formula (for a pole of order m at $z_0$): $Res(f;z_0)=\lim_{z \to z_0} \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z))$

Answer 1

If a function $f$ has Laurent Series expansion $$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$$

then residue of $f$ at $z_0$ is $a_{-1}$ where $a_{-1}$ is the coefficient of the term $(z-z_0)^{-1}$.

So, if there is no term $(z-z_0)^{-1}$ in $f'$, the coefficient, say $a'_{-1}$, is $0$ therefore the residue is $0$.

Answer 2

The definition of the residue is the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion.

Or, if you must use the Residue Theorem: If $\gamma:[a,b]\to\mathbb C$ and $f$ is differentiable on $\gamma$ the Fundamental Theorem of Calculus shows that $$\int_\gamma f'(z)\,dx=f(\gamma(b))-f(\gamma(a)).$$So we always have $\int_\gamma f'=0$ if $\gamma $ is a closed curve.