Consider

Now the numerator has 4 zeroes. And the denominator has 6 zeroes? And and have triple zeroes in the denominator?

Thus has a double pole at ?

Is this correct?

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- Apr 11th 2009, 03:10 PMmanjohn12pole
Consider

Now the numerator has 4 zeroes. And the denominator has 6 zeroes? And and have triple zeroes in the denominator?

Thus has a double pole at ?

Is this correct? - Apr 16th 2009, 11:19 AMMoo
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...