Residues at simple poles

Dec 2009
3
0
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
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.
 
Dec 2009
3
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
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
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.