Thread: 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

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

Because near

CB

And that is because .

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

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



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