Thread: Residues at simple poles

Hi folks,

Having real problems with a complex analysis homework. can the residue of a complex function at a simple pole be 0?

ie for g(z)= sin(z)/z i think the pole is just 0 of order one which is simple, but by calculating the residue i get 0..am i doing something wrong?

I'm using the formula if c is the simple pole and g(z)= f(z)/h(z) then the residue at c of g(z) = f(c)/h'(c)..is that okay?

Thanks for any help

Donsie

Originally Posted by Donsie

Hi folks,

Having real problems with a complex analysis homework. can the residue of a complex function at a simple pole be 0?

ie for g(z)= sin(z)/z i think the pole is just 0 of order one which is simple, but by calculating the residue i get 0..am i doing something wrong?

I'm using the formula if c is the simple pole and g(z)= f(z)/h(z) then the residue at c of g(z) = f(c)/h'(c)..is that okay?

Thanks for any help

Donsie

There is no pole at all at z= 0 for , just a "removable discontinuity". Remember, from Calculus I, that . "Remove" the discontinuity by defining g(0)= 1.

The Taylor's series for sin(z) is .

So , an analytic function. Again, that has the value "1" at z= 0.

thank you!

So to make sure i have it right..

the function f(z)= 1/1+z^3 has a simple pole at z= -1. correct?

Then to get the residue at this pole can i use the formula res(f,c) where c is the simple pole = g(c)/h'(c) where g(c) is just 1, h'(z)= 3z^2 so h'(c)=3. so the residue is just 1/3? is that also correct?

Thank you for your patience!!

Donsie

Originally Posted by Donsie

thank you!

So to make sure i have it right..

the function f(z)= 1/1+z^3 has a simple pole at z= -1. correct?

Then to get the residue at this pole can i use the formula res(f,c) where c is the simple pole = g(c)/h'(c) where g(c) is just 1, h'(z)= 3z^2 so h'(c)=3. so the residue is just 1/3? is that also correct?

Thank you for your patience!!

Donsie

Yes. But note that z = -1 is not the only simple pole. There are simple poles at values of z such that z^3 + 1 = 0.