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Im struggling with the concept of finding poles and then indicating their order..

The following past exam questions asks to find the poles and indicate their order

$f(z)=\dfrac{z^2-3z+2}{(z-1)^2(z-3)^2}$

My workings:

$f(z)=\dfrac{(z-1)(z-2)}{(z-1)^2(z-3)^2}$

$f(z)=\dfrac{z-2}{(z-1)(z-3)^2}$

I think this means I have poles at +3 and +2 with an order of 2.. Im not sure if this is right..

Another example I have come across from a past exam paper is :

$g(z)=\dfrac{1}{((z-i)(z-4i))^2}$

$g(z)=\dfrac{1}{(z^2-5iz-4)^2}$

but im not sure where to go from here because multiplying that out would involve $z^4$ etc

Mathsstudent147
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1 Answers1

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You are not correct about the poles of $f$. There are poles at $z=1$ and $z=3$.

It is clear f is not holomorphic at $z=1$. If $z=1$ is a pole of order 1, then $$ lim_{z \rightarrow 1}(z-1)f(z)=C $$ for some constant $C$. If we try with this $f$ we get $$ lim_{z \rightarrow 1}(z-1)\frac{z-2}{(z-1)(z-3)^2}=lim_{z \rightarrow 1}\frac{z-2}{(z-3)^2} = \frac{-1}{4} $$ So it is a pole of order 1. Using a similar argument, the pole at $z=3$ is of order 2 and there are no other poles.

Amy
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