# pole

• Apr 11th 2009, 03:10 PM
manjohn12
pole
Consider $\displaystyle f(z) = \frac{e^{z}(1- \cos(z))^{2}}{\sin (z)^{3} z^{3}}$

Now the numerator has 4 zeroes. And the denominator has 6 zeroes? And $\displaystyle \sin z$ and $\displaystyle z$ have triple zeroes in the denominator?

Thus $\displaystyle f$ has a double pole at $\displaystyle 0$?

Is this correct?
• Apr 16th 2009, 11:19 AM
Moo
Quote:

Originally Posted by manjohn12
Consider $\displaystyle f(z) = \frac{e^{z}(1- \cos(z))^{2}}{\sin (z)^{3} z^{3}}$

Now the numerator has 4 zeroes. And the denominator has 6 zeroes? And $\displaystyle \sin z$ and $\displaystyle z$ have triple zeroes in the denominator?

Thus $\displaystyle f$ has a double pole at $\displaystyle 0$?

Is this correct?

I don't know what you mean by a "zero", but it is true that 0 is a zero of the numerator with multiplicity 4, and multiplicity 6 in the denominator.
If it is indeed a pole (you have to prove it by finding the Laurent series of this function in a neighbourhood of 0), it is likely to be a double pole...