# Laurent expansion of $f(z)=ize^z$

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!

• Well... $f$ is a holomorphic function, so sure, it has a Laurent expansion, but this is nothing other than the Taylor expansion, so what you did is right (you could have even stopped at $f(z)=\sum_{k=0}^{\infty}\frac{i}{k!}z^{k+1}$, there's no need to change the index). From this, what is the residue? It should be trivial to read off the answer. Apr 23, 2021 at 0:06
• @peek-a-boo Im sorry to ask again but as i said im new at this and kinda confused i know that to find the residue point you need to find $a_{-1}$ so in this case i substitute n with -1? Does it even make sense to have the factorial of a negative number? Apr 23, 2021 at 0:16
• nonono you're going way off track. Can you write out what are the first few terms of the series? Atleast write out the first 2 terms Apr 23, 2021 at 0:16
• @peek-a-boo So starting from 1 right? we get : $iz$ +$iz^2$?? Apr 23, 2021 at 0:18

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.

• So the residue is always the index of ${1/z}$? and in cases i cannot "compute" it it is 0? Right? Apr 23, 2021 at 0:26
• @AniLici yes, the residue (at the origin) is by (one of many equivalent) definition the coefficient of $\frac{1}{z}$ in the Laurent expansion. If such a term does not appear, then the residue is $0$. For example, the function $f(z)=\frac{1}{z^2}$ is already in "Laurent form", and this has zero residue at $0$ because there is no $\frac{1}{z}$ term. Another example is $f(z)=\left(\frac{\cos z}{z}\right)^2=\frac{1}{z^2}- 1 +\dots$, so again this has $0$ residue at the origin. Apr 23, 2021 at 0:31
• God bless you man! You just made things so clear! One more question please! Iff for example $z_0$ is not 0 but another number for example like -1! What would change? Apr 23, 2021 at 0:41
• @AniLici for the original function in your question, nothing changes, because like I said, your function is holomorphic at every point, so the Laurent expansion about any point is simply the Taylor expansion. So, there will never be $\frac{1}{z-z_0}$ terms, only $(z-z_0)^k$ where $k\geq 0$. FOr the examples I made up in my comment, stuff might change (it probably will) because the Laurent expansion isn't as obvious anymore, some extra calculations need to be done... which I don't feel like doing now :) Apr 23, 2021 at 0:49
• you dont need to! You already have helped me too much and sorry to be a bother! Thank you so much! Apr 23, 2021 at 0:53