# Problem with pole

• Feb 22nd 2010, 04:10 AM
Mohit
Problem with pole
Consider the function: {cot(z)*coth(z)} /z^3

does it have a pole at z=3 ? If so, can you prove it ?
• Feb 22nd 2010, 08:23 AM
HallsofIvy
Quote:

Originally Posted by Mohit
Consider the function: {cot(z)*coth(z)} /z^3

does it have a pole at z=3 ? If so, can you prove it ?

Well, I suppose one could used the definition of "pole"!

$z_0$ is a "pole of order n" for f(z) if and only if the Laurent series for f(z) around $z_0$ has [tex]z^{-n} with non-zero coefficient but no power lower than -n.

But the crucial point here is that cot(z) and coth(z) are analytic at z= 3 while $1/z^3$ is a rational function with non-zero denominator at z= 3 and so also analytic there. All three functions have Taylor's series (Laurent series with no negative powers) about z= 3 and so there product has a Taylor's series there.

A more interesting problem would be to show that $f(z)= cot(z)coth(z)/z^3$ has a pole of order 3 at z= 0.
• Feb 22nd 2010, 12:46 PM
chisigma
Quote:

Originally Posted by HallsofIvy
... a more interesting problem would be to show that $f(z)= cot(z)coth(z)/z^3$ has a pole of order 3 at z= 0...

... are You sure?(Wink) ...

Kind regards

$\chi$ $\sigma$
• Feb 23rd 2010, 04:40 AM
HallsofIvy
:o Assuming, of course, that neither tan(x) nor tanh(x) is 0 at x= 0!!!