If $f$ is even function, then both main part and regular part of its Laurent expansion are even?

Question

So,I have given a function which is analytic in an annulus, so I can do the Laurent expansion. If I know that $f$ is even (odd) ,how can I prove that then both main part and regular part in Laurent expansion are even (odd) functions? I proved that if $f$ is even than the Laurent expansion has only even degrees of z , hence only even coefficients in expansion. But I'm not sure if this helps in this or how can I actually check the parity of main and regular part?

Thank you in advance.

Answer 1

Hint $$a_n=\frac{1}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz $$

Do the substitution $u=-z$.

Note that after the substitution you get $$a_n=\frac{1}{2 \pi i} \int_{|z|=R} \frac{f(-u)}{(-u)^{n+1}} d(-u)=(-1)^{n+2}\frac{1}{2 \pi i} \int_{|z|=R} \frac{f(-u)}{u^{n+1}} du $$

Now, if $f$ is even, since $f(-u)=f(u)$ you get $$a_{n}=(-1)^{n+2}a_n \forall n$$

Now, if $f$ is odd, since $f(-u)=-f(u)$ you get $$a_{n}=(-1)^{n+3}a_n \forall n$$

P.S. If all the odd terms are zero then the regular part is $$g(z)=\sum_{n=0}^\infty a_n z^n=\sum_{k=0}^\infty a_{2k}z^{2k}$$

What is $g(-z)$?

Same way, the main part is $$h(z)=\sum_{n=1}^\infty a_{-n} z^{-n}=\sum_{k=1}^\infty a_{-2k}z^{-2k}$$

What is $h(-z)$?

You can then do the same for odd functions. Or you can get the odd function for free by observing that $f$ is odd if and only if $zf(z)$ is even.

Answer 2

If $f$ is even and $f(z)=\sum_{n=-\infty}^\infty a_nz^n$, then $f(z)-f(-z)=0$. But$$0=f(z)-f(-z)=\sum_{n\in\mathbb Z,\ n\text{ odd}}2a_nz^n.$$So, for each odd integer $n$, $a_n=0$, and therefore$$f(z)=\sum_{n\in\mathbb Z,\,n\text{ even}}a_nz^n.$$