# How to compute $f(z) = \sum_0^{\infty} (1+2i+(2+i)(-1)^k)^{-k}z^k$

How to compute this serie :

$$f(z) = \sum_0^{\infty} (1+2i+(2+i)(-1)^k)^{-k}z^k$$

The serie is convergent if $|z| < \sqrt{2}$

I can find that $$f(z) = \sum_0^{\infty} 3^{-2k}(1+i)^{-2k}z^{2k} + \sum_0^{\infty} (-1+i)^{-2k-1}z^{2k+1}$$

I have $$\sum_0^{\infty} 3^{-2k}(1+i)^{-2k}z^{2k} = \frac{18}{18+iz^2}$$

But I don't know how to deal with the $\sum_0^{\infty} (-1+i)^{-2k-1}z^{2k+1}$ part.

As $\displaystyle\frac1{-1+i}=\frac{-1-i}{(-1)^2-i^2}=\frac{-i-i}2$
$$(-1+i)^{-2(k+1)}z^{2k+1}=\left(\frac z{-1+i}\right)^{2k+1}=\left(\frac {-z(1+i)}2\right)^{2k+1}=-\left(\frac {z(1+i)}2\right)^{2k+1}$$
As $$(-1+i)^{-2k-1}z^{2k+1}=\frac{z^{2k+1}}{(-1+i)^{2k+1}}=\left(\frac{z}{-1+i}\right)^{2k+1}= \left(\frac{z}{-1+i}\right) \left[\left(\frac{z}{-1+i}\right)^2\right]^k$$ then $$\sum_{k=0}^{+\infty}(-1+i)^{-2k-1}z^{2k+1}= \left(\frac{z}{-1+i}\right) \sum_{k=0}^{+\infty}\left[\left(\frac{z}{-1+i}\right)^2\right]^k$$ and the last one is a geometric series, and it converges as $\left|\frac{z}{-1+i}\right|<1$ (in fact we suppose $|z|<\sqrt2$). So the sum is $$\left(\frac{z}{-1+i}\right)\frac{1}{1-\left(\frac{z}{-1+i}\right)^2}= \left(\frac{z}{-1+i}\right)\frac{(-1+i)^2}{(-1+i)^2-z^2}=\frac{z(-1+i)}{-2i-z^2}$$
• Yes .. except you forgot to square the $z/(-1+i)$, and also forgot to multiply by the $z/(-1+i)$ you took outside the summation. Commented Apr 13, 2014 at 14:38
• Thanks, I had a weird form and I did not see the $\left|\frac{z}{-1+i}\right|<1$ Commented Apr 13, 2014 at 15:25