Even and odd functions in Laurent decomposition

Question

I have the following problem and have no idea how to approach it, could anyone give me any hint about it? Thanks!

Suppose that $f(z) = f_0(z) + f_1(z)$ is the Laurent decomposition of an analytic function $f(z)$ on the annulus $\{A < |z| < B \}$. Show that if $f(z) $ is an even function, then $f_0(z)$ and $f_1(z)$ are even functions.

Answer 1

$$ f_0(-z) + f_1(-z) = f(-z) = f(z) = f_0(z) + f_1(z) $$ in $\{A < |z| < B \}$, and the Laurent decomposition (with the normalization $\lim_{z \to \infty} f_1(z) = 0$) is unique:

Define $h : \Bbb C \to \Bbb C $ by $$ h(z) = \begin{cases} f_0(-z) - f_0(z) & \text{ for } |z| < B \\ f_1(z) - f_1(-z) & \text{ for } |z| > A \\ \end{cases} $$ and verify that

• $h$ is well-defined,
• $h$ is an entire function, and finally,
• $h$ is identically zero.

Answer 2

Continuing with Martin R's answer, we have that $\lim_{z \to \infty} f_1(z) = 0$, so $f_1$ is bounded.

Also, since $f_0$ is analytic, and hence continuous, on the compact set $|z| < B$, $f_0$ is bounded. Therefore, $h(z)$ is bounded and analytic on the entire plane. Liouville's theorem implies that $h(z)$ is identically zero.

Therefore, both $f_0$ and $f_1$ are even.