Find isolated singular points of $f(z)$ and

Question

Find isolated singular Points of $f(z)$, then explain what the type of this point :

1) $f(z)={{z}\over {e^{z-1}}}$

2) $f(z)={{z+1}\over {z^2 -2z}}$

3) $f(z)={1-e^{2z}\over z^4}$

My try:

3) $f(z)={1-e^{2z}\over z^4}$

$f(z)={1\over z^4}[1-(1+2z+{{(2z)^2}\over 2!}+... )]$

$={-1\over z^4}(2z+{{(2z)^2}\over 2!}+... )$

$=-({2\over {z^3}} + {2\over {z^2}} + {{2^3}\over z}+{2\over 3}+...)$

then the isolated singular point is $0$, and it's type (pole - order 3)

true ?

and what about problems 1 and 2 ?

"note: I don't speak english well :("

Answer 1

Assuming that you are not using $z=\infty$, then:

3 should be correct.

1: Use $ze^{1-z}$, and then expansion is easy: $z+\frac{z(1-z)}{1!}+\frac{z(1-z)^2}{2!}+...$

2: It's a rational function so it's just the zeroes of the denominator (z=0, 2)

Considering the point at infinity, you should watch out on the third one.