# Finding the sum of a given Laurent series

A Laurent series around $0$ has $c_k=(-2)^k \frac {1} {(2k)!},k>0$; $c_0=0$; $c_k=(-1)^k\frac {1} {k},k<0$, I found the annulus of convergence to be $|z|>1$, How to find the sum function of this series? And what are the singular points on the boundary of annulus?

I'll just give some hints: The positive part of the series (meaning terms with $k\ge0$) looks very much like the series for the cosine function. I suggest $\cos(\sqrt{2z})-1$ as a likely candidate. For the negative part, think logarithms. Now add the two together. And it should converge for $\lvert z\rvert>1$.

The annulus of convergence is actually $\{z\;;\;|z|>1\}$.

On one hand, we have $$\sum_{k\geq 1}(-2)^k\frac{z^k}{(2k)!}=\sum_{k\geq 1}\frac{(-2z)^k}{(2k)!}=-1+\sum_{k\geq 0}(-1)^k\frac{(2z)^k}{(2k)!}=-1+\cos(\sqrt{2z})$$ where $\sqrt{2z}$ is the multi-valued square root function. This makes sense because $\cos$ even, and takes the same value on both square roots. And this converges for every $z$.

On the other hand, we get $$\sum_{k\geq 1} (-1)^{-k}\frac{z^{-k}}{-k}=\sum_{k\geq 1} (-1)^{k+1}\frac{(1/z)^k}{k}=\log\left(1+\frac{1}{z} \right).$$ This converges for $|1/z|<1$, that is $|z|>1$, and it diverges of $|z|<1$. So your Laurent series adds up to $$-1+\cos (\sqrt{2z})+\log\left(1+\frac{1}{z} \right)\qquad \forall |z|>1.$$

Regarding singular points on the boundary $|z|=1$, it amounts to the singular points of $\log (1+z)$ on the unit circle.

Note: if you like $\cosh$ better, you can replace $\cos (\sqrt{2z})$ by $\cosh (\sqrt{-2z})$. Also, what you wrote is not correct. The coefficients $c_k$ should not feature $z^k$.

• I think you want to reconsider the hyperbolic cosine. Commented Apr 2, 2013 at 15:33
• @HaraldHanche-Olsen I think this agrees with your approach as $\cosh (iz)=\cos z$ and vice versa. Commented Apr 2, 2013 at 15:36
• You're right, of course. Duh. Commented Apr 2, 2013 at 15:39
• @HaraldHanche-Olsen But $\cos$ is used more often, so I'll go like you. Commented Apr 2, 2013 at 15:42
• Thank you, i have made the corrections Commented Apr 2, 2013 at 16:55