How to prove that this two functions have simple poles:

$\displaystyle f(z)=\frac{cos(z)}{1-2sin(z)}$ at $\displaystyle z=\pi/6$

$\displaystyle g(z)=\frac{senh(z)}{cosh(z)-1}$ at $\displaystyle z=2ki\pi$ where k is an integer

Regards.

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- May 12th 2011, 05:01 AMhurzProving the simple poles
How to prove that this two functions have simple poles:

$\displaystyle f(z)=\frac{cos(z)}{1-2sin(z)}$ at $\displaystyle z=\pi/6$

$\displaystyle g(z)=\frac{senh(z)}{cosh(z)-1}$ at $\displaystyle z=2ki\pi$ where k is an integer

Regards. - May 12th 2011, 05:04 AMHallsofIvy
Well, what is the

**definition**of "simple pole"? - May 12th 2011, 05:12 AMhurz
you have a simple pole when the maximum power of 1/z you have is 1 in the laurent series of the function

- May 12th 2011, 09:40 AMjamesdt
I'm not sure I understood your definition of a 'simple pole'. Is not a pole commonly defined as the values of $\displaystyle z$ such that the denominator is zero? And perhaps by simple, you mean that the pole has multiplicity one?

For example, the first function $\displaystyle f(z)$ would have a pole when $\displaystyle 1-2\sin(z)=0$, or equivalently $\displaystyle \sin(z)=\frac{1}{2}$. Then $\displaystyle z=\frac{\pi}{6}$ certainly is a pole, but since $\displaystyle \sin(z)$ is periodic, there most certainly are other poles also.