I started complex analysis a couple of weeks ago and we stated the Laurent expansion but i cant warp my head around them ! There was this exercise that i found on the practice exercise session
Determine the Laurent expansion of the function $f(z)=ize^z$ around the origin, characterize such point and compute the residue.
What i have done for now is that i expanded the expression into $f(z)=\sum_{k=0}^{\infty} \frac{i}{k!}z^{k+1}$
and then i substituted k for n getting : $f(z)=\sum_{n=1}^{\infty} \frac{i}{(n-1)!}z^{n}$
but im not able to continue further!
Ok good, you have the first few terms of the series, and of course, the other terms are all higher order terms. Now, don't think in term of formulas "$a_{-1}$ is the residue" because obviously the formulas are what's confusing you when you tried to change the index. Think of the definition in words: "the residue is the coefficient of $\frac{1}{z}$ in the Laurent expansion". Ok, so as you have just calculated, the Laurent expansion is really a Taylor expansion, so there are no inverse powers of $z$. i.e there is no $\frac{1}{z}$ or $\frac{1}{z^2}$ or any of that stuff. This is the same as saying the coefficient of $\frac{1}{z}$ in the Laurent expansion is $0$. Therefore, the residue is...
Another remark I feel compelled to add. Your series is $\sum_{k=0}^{\infty}\frac{i}{k!}z^{k+1}=iz + iz^2 + \dots$ So no amount of index changing can suddenly cause a $\frac{1}{z}$ term to appear. Keep in mind that $\sum$ notation is just a convenient way of writing things down, it can't magically create or destroy things which aren't there.