Consider the function: {cot(z)*coth(z)} /z^3

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

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- Feb 22nd 2010, 04:10 AMMohitProblem 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 AMHallsofIvy
Well, I suppose one could used the

**definition**of "pole"!

$\displaystyle z_0$ is a "pole of order n" for f(z) if and only if the Laurent series for f(z) around $\displaystyle z_0$ 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 $\displaystyle 1/z^3$ 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 $\displaystyle f(z)= cot(z)coth(z)/z^3$ has a pole of order 3 at z= 0. - Feb 22nd 2010, 12:46 PMchisigma
- Feb 23rd 2010, 04:40 AMHallsofIvy
:o Assuming, of course, that neither tan(x) nor tanh(x) is 0 at x= 0!!!