# Thread: Residues at simple poles

1. ## 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

2. 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 $\displaystyle \frac{sin(z)}{z}$, just a "removable discontinuity". Remember, from Calculus I, that $\displaystyle \lim_{z\to\infty}\frac{sin(z)}{z}= 1$. "Remove" the discontinuity by defining g(0)= 1.

The Taylor's series for sin(z) is $\displaystyle z- \frac{1}{6} z^3+ \frac{1}{5!}z^5+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}z^{2n+1}+ \cdot\cdot\cdot$.

So $\displaystyle \frac{sin(z)}{z}= 1- \frac{1}{6}z^2+ \frac{1}{5!}z^4+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}z^{2n}+ \cdot\cdot\cdot$, an analytic function. Again, that has the value "1" at z= 0.

3. ## ..

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?

Donsie

4. 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?