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.

  • $\begingroup$ 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 $|z-1| > 1$. $\endgroup$
    – copper.hat
    Nov 12, 2016 at 0:58
  • $\begingroup$ @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? $\endgroup$ Nov 12, 2016 at 1:01
  • $\begingroup$ 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. $\endgroup$ Nov 12, 2016 at 2:14
  • $\begingroup$ What happens with the cosine on top? Do I just leave that out front as a multiplier? $\endgroup$
    – Spuds
    Nov 12, 2016 at 3:52

1 Answer 1


Here are some hints:

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 } $


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