# Taylor and Laurent Series of Complex Functions

I'm dealing with the function:

$w=\frac{cos(z)}{z^2}$ with $z_0=1$.

I need to find the Taylor and Laurent series expansions about $z_0$ and find their regions on convergence. I'm pretty confused on how to do these, we haven't done anything comparable in class.

Any help would be appreciated.

• Note that the function is analytic except at $z=0$. Hence it will have a Taylor series for $|z-1| <1$ and a Laurent expansion for $|z-1| > 1$. Nov 12, 2016 at 0:58
• @copper.hat Won't the Laurent series equal the Taylor series...? (Sorry, I'm kind of new to this) Oh wait, because of poles, so that determines the areas where each converges, right? Nov 12, 2016 at 1:01
• Hm, no, I think that according to how coefficients of a Laurent series are defined, all the negative powers become $0$ and we are left with the Taylor expansion. Nov 12, 2016 at 2:14
• What happens with the cosine on top? Do I just leave that out front as a multiplier? Nov 12, 2016 at 3:52

Note that for $z \neq 0$, $f(z) = {1 \over z^2} - 2 {1 \over z^4} + \sum_{n=0}^\infty (-1)^n {z^n \over (n+2)!}$.
Let $r(z) = \sum_{n=0}^\infty (-1)^n {z^n \over (n+2)!}$, note that $r$ is entire, and we can write $r(z) = \sum_{n=0}^\infty {r^{(n)}(1) \over n!} (z-1)^n$, with $r^{(n)}(1) = \sum_{n=0}^\infty (-1)^n n (n-1)\cdots (n-k+1){1 \over (n+2)!}$
For $|z-1| <1$ we can write ${1 \over z^k} = ( 1+(z-1))^{-k} = \sum_{n=0}^\infty \binom{-k}{n}(z-1)^n$.
For $|z-1| >1$ we can write ${1 \over z^k} = {1 \over (1+(z-1))^{k} } = {1 \over (z-1)^k} {1 \over (1+{1 \over (z-1)) } )^{k} } = {1 \over (z-1)^k} \sum_{n=0}^\infty \binom{-k}{n}{1 \over (z-1)^n }$