# Poles or Removable Singularities

• Aug 29th 2009, 01:14 AM
Chris0724
Poles or Removable Singularities
hi,

Given the equation below, it a pole of order 5 when z = 0

f(z) = sin(z)/z^6

Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?

Thanks (Talking)
• Aug 29th 2009, 02:23 AM
CaptainBlack
Quote:

Originally Posted by Chris0724
hi,

Given the equation below, it a pole of order 5 when z = 0

f(z) = sin(z)/z^6

Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?

Thanks (Talking)

Because $\frac{\sin(z)}{z^6} \approx z^{-5}$ near $z=0$

CB
• Aug 29th 2009, 03:58 AM
HallsofIvy
And that is because $\lim_{x\to 0}\frac{sin(x)}{x}= 1$.
• Aug 29th 2009, 09:25 AM
shawsend
Quote:

Originally Posted by Chris0724
hi,

Given the equation below, it a pole of order 5 when z = 0

f(z) = sin(z)/z^6

Question--> I can't seem to get how to achieve the pole. The answer was given by the prof so i assume it's right.

Why is it not RS when we apply limit z => 0 ? isn't the whole equation set out to be 0 , then using L'hopital rule to find the answer?

Thanks (Talking)

Also, just do long division to calculate the Laurent series:

$\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots}{z^6}=\frac{1 }{z^5}-\frac{1}{6z^3}+\cdots$

The answer tells you immediately it's a pole of order five.