If $f(z)$ is even then residue at $z=z_0$ is zero?

Question

There is a theorem in my text book which says that:

Theorem:If $f$ has an isolated singularity at $z_0$ and if $f$ is even in $z-z_0$ i.e $f(z-z_0)=f(-(z-z_0))$ then Res$[f(z),z_0]=0$ [Theorem 8.60, page 290,S ponnusami]

My doubt: let $f(z)$ be even function then $f(z)=f(-z)$ for all $z$ in domain of f and hence we have $f(w)=f(-w)$ where $w=z-z_0$ implying that $f$ is even in $z-z_0$ and hence from theorem for any even function $f$ Res$[f(z),z_0]=0$ at isolated singularity $z_0$ which I think wrong.For counter example see the picture added

Answer 1

I think what you've gotten wrong is the theorem itself. When faced with a statement you doubt, I recommend looking for easy examples before you try and prove it. Here, I can think of the function:

$$f(z)= \frac{1}{z^2-1}$$ It is clearly even, since $ f(z)=f(-z) $. If we calculate the residue of f in $z_0=1$, we get: $$\lim_{z\rightarrow1}{f(z)·(z-1)}=\lim_{z\rightarrow1}{\frac{1}{z+1}}=\frac{1}{2}\neq0 $$ If a function is even, it will have a singularity at $z_0$ if and only if it has it at $-z_0$, and the residues of $f$ will be the same.

Answer 2

The theorem requires that the function be even about the point $z_0$. The function $\frac{1}{z \sin z}$ is even about $z=0$, but it is not even about $z=\pi$. Indeed, the residue at $z=0$ is in fact $0$, but not the residue at $z=\pi$.