Entire Function, Taylor Series Suppose $f(z)$ is entire and for all $z \in \mathbb{C}$, $f(z) = f(-z)$. Let $\displaystyle{g(z) = f \left(iz-\frac{i}{z} \right)}$, for $z \neq 0$. Prove that $\displaystyle{g(z) = c_0 + \sum_{k=1}^\infty c_k \left( z^{2k} + \frac{1}{z^{2k}} \right), \; z \neq 0}$ where $\displaystyle{c_k = \int_0^\pi f(2 \sin t) \cos 2kt \; dt}$ for $k \geq 0$.
My attempt:
$f(z)$ is entire so it has a Taylor series representation about the origin $\displaystyle{f(z) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} z^k}$.
For all $z \in \mathbb{C}$, $f(z) = f(-z)$. Then this implies that $f^{(k)}(0) = 0$ for every odd $k \geq 0$.
So we may write $\displaystyle{f(z) = \sum_{k=0}^\infty \frac{f^{(2k)}(0)}{(2k)!} z^{2k}}.$ 
How may I continue my proof?
 A: We have
$$
g(z)=\sum_{n=-\infty}^\infty a_nz^n \quad \forall z \in \mathbb{C},
$$
with
$$
a_n=\frac{1}{2\pi i}\int_{|z|=1}\frac{g(z)}{z^{n+1}}\,dz.
$$
Notice that
$$
a_n=\frac{1}{2\pi}\int_0^{2\pi} e^{-int}g(e^{it})\,dt=\frac{1}{2\pi}\int_0^{2\pi}e^{-int}f(-2\sin t)\,dt=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-int}f(2\sin t)\,dt,
$$
therefore
\begin{eqnarray}
\Im a_n&=&-\frac{1}{2\pi}\int_{-\pi}^\pi\sin(nt)f(2\sin t)\,dt=0,\\
a_n&=&\Re a_n=\frac{1}{2\pi}\int_{-\pi}^\pi \cos(nt)f(2\sin t)\,dt=\frac{1}{\pi}\int_0^\pi\cos(nt)f(2\sin t)\,dt.
\end{eqnarray}
Also
\begin{eqnarray}
a_{2n+1}&=&\frac{1}{2\pi}\int_0^{2\pi}\cos((2n+1)t)f(2\sin t)\,dt\\
&=&\frac{1}{2\pi}\int_0^\pi \cos((2n+1)t)f(2\sin t)\,dt+\frac{1}{2\pi}\int_\pi^{2\pi} \cos((2n+1)t)f(2\sin t)\,dt\\
&=&\frac{1}{2\pi}\int_0^\pi \cos((2n+1)t)f(2\sin t)\,dt-\frac{1}{2\pi}\int_0^\pi \cos((2n+1)t)f(2\sin t)\,dt\\
&=&0.
\end{eqnarray}
Setting 
$$
c_n:=a_{2n}=\frac1\pi\int_0^\pi \cos(2nt)f(2\sin t)\,dt,
$$
we get for every $z \in \mathbb{C}\setminus\{0\}$:
$$
g(z)=a_0+\sum_{n=1}^\infty\left(a_{2n}z^{2n}+\frac{a_{-2n}}{z^{2n}}\right)=c_0+\sum_{n=1}^\infty c_n\left(z^{2n}+\frac{1}{z^{2n}}\right).
$$
