Finding poles and their order for f(z) = z³/(2z+1)³

[FanofAFan]

Homework Statement

find the pole and it's order for f(z) = $$\frac{z^3}{(2z+1)^3}$$

Homework Equations

$$\frac{1}{(2z+1)^3}$$ as $$\frac{1}{8(z+1/2)^3}$$

The Attempt at a Solution

poles at z = -1/2, $$\frac{e^(i*\pi)}{2}$$, and $$\frac{e^(-i*\pi)}{2}$$

 

[ideasrule]

e^(iπ) and e^(-iπ) are both -1, so they're not poles. I think you're overthinking the problem. For any function $$f(z) = \frac{g(z)}{(z-a)^n}$$, "a" is a pole with order n. If the denominator was (z-a)^m*(z-b)^n, then both a and b would be poles, with orders m and n.

 

[FanofAFan]

Ok, I know -1/2 had to be a pole but I meant e^(ipi)/2 which is also -1/2
and also when i try to find the residue at that point I end up getting constant/zero...

ie res at -1/2 = (-1/2)^3/24(-1/2+1/2)^2 is that wrong?