# 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?

• I would start from the other end. $g$ is holomorphic on $\mathbb{C}\setminus \{0\}$. Thus $g$ has a Laurent series expansion about $0$, $g(z) = \sum_{k=-\infty}^\infty b_kz^k$. Now use the properties of $f$ to see that a) $b_{-k} = b_k$ for all $k$, b) $b_{2k+1} = 0$ for all $k$, and c) the $c_{k} = b_{2k}$ have the given integral representation. – Daniel Fischer Nov 7 '13 at 16:37
• It should be $c_k=\frac1\pi\int_0^\pi (2\sin t)\cos(2kt)\,dt$ – Mercy King Nov 7 '13 at 18:29

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 $z \in \mathbb{C}\setminus\{0\}$: $$g(z)=a_0+\sum_{n=1}^\infty\left(a_{2n}z^{2n}+\frac{a_{-2n}}{z^{2n}}\right)=c_0+\sum_{n=1}^\infty c_n\left(z^{2n}+\frac{1}{z^{2n}}\right).$$