# Complex analysis problem with laurent series and singular points

I'm trying to figure out problem b from this set of homework questions:

Find the Laurent series expansions of $$\textrm{(a) } \frac{e^z}{z^4},\qquad \textrm{(b) } ze^{1/z},\qquad \textrm{(c) } \frac{1}{z(z-1)}$$

The expansion is taken around zero.

Currently it's my understanding that you take a Taylor expansion of a part of the function that is analytic and then multiply by the bit that remains. However for question b my lecturer has taken a Taylor expansion for $e^{1/z}$. My confusion is that isn't $z = 0$ for $e^{1/z}$ a singular point and therefore he can't take a Taylor expansion of it and simply multiple by $z$ to get his answer?

Any help would be appreciated

• You seriously need to typeset, now here at MSE, those questions from the link. Tha link, after all, can easily disappear over time. – zhw. Feb 22 '17 at 22:41
• So how do I show the problem as you did without a link? – Sebastian TG Feb 23 '17 at 10:00
• oh I see how you have done that – Sebastian TG Feb 23 '17 at 10:01

$e^{\frac{1}{z}}$ does have an essential singularity at zero, it doesn't have a Taylor series, but it does have a Laurent expansion. $$e^{\frac{1}{z}}=\cdots \frac{1}{2!z^2}+\frac{1}{z} +1$$
• I took the usual taylor series for $e^z$ and substituted $\frac{1}{z}$. – Rene Schipperus Feb 22 '17 at 22:29
You know that for all $w \in \mathbb C$ you have $$e^w=1+\frac{1}{1!}w+\frac{1}{2!}w^2+...$$
Now, if $z \in \mathbb C$ is non zero, setting $w=\frac{1}{z}$ you get $$e^{\frac{1}{z}}= 1+\frac{1}{1!z}+\frac{1}{2!z^2}+...$$