# Finding the analytic continuation of $f(z)$

Given

$$f(z)=\sum_{n=0}^{\infty} (-1)^nz^{2n}$$

for $|z|<1$

I'm tasked with finding the analytic continuation of $f(z)$ to the entire z-plane. Also, are there any points where the continuation in not analytic?

Analytic continuation is new to me so I'm unsure what how to approach the problem. My instinct is to use the real axis, where $f(z)=f(x)$, to relate the above series to a known real series where I know the explicit expression for $f(x)$. Would this be a valid approach, or should I proceed by different means?

• What is $f(z)?$ And why do you think that equality holds for $|z|<1?$ – zhw. Mar 7 '16 at 21:23
• My mistake - it's the series above, which I've now corrected. In regards to the equality, $f(z)=f(x)$ for $|x|<1$ as well? – Tunk Mar 7 '16 at 21:27

Note that for $|z|<1$ the series converges and we have
$$\sum_{n=0}^\infty (-1)^nz^{2n}=\frac{1}{1+z^2}$$
for $|z|<1$. Note that the function $f(z)$ given by
$$f(z)=\frac{1}{z^2+1}$$
is analytic everywhere except at the simple poles $z=\pm i$. It is the analytic continuation of the series for $z\ne \pm i$.
That sum is a geometric series $\sum_{n=0}^\infty a_n$, with general term $a_n =(-z^2)^n$. Use the formula for sum of convergent geometric series to get an analytic (or at least meromorphic) expression that is valid in most of the plane.