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

I know the zeroes are z=0 with order 2 and z=1 with order 3. I know how to find residue of simple poles but i am confused about finding residues of multiple poles?

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  • $\begingroup$ $z=0$ and $z=1$ are not zeroes, they are poles. $\endgroup$ – GEdgar Oct 22 at 17:01
  • $\begingroup$ sorry i meant poles $\endgroup$ – bow123 Oct 22 at 17:09
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The residue of a order $n$ pole can be calculated as $$b_1={1\over (n-1)!}\lim _{z\to z_0}{d^{n-1}\over dx^{n-1}}z^nf(z)$$This is because for an order $n$ pole $z_0$ we have $$f(z)=\sum_{k=1}^n{b_k\over (z-z_0)^k}+g(z)$$therefore$$(z-z_0)^nf(z)=\sum_{k=1}^n{b_k (z-z_0)^{n-k}}+(z-z_0)^ng(z)$$by $n-1$ times differentiating and tending $z$ to $z_0$, we obtain the result.

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  • $\begingroup$ Can you show how this would work for either z=0 or z=1? $\endgroup$ – bow123 Oct 22 at 17:02
  • $\begingroup$ Sure. I added some more details. $\endgroup$ – Mostafa Ayaz Oct 22 at 17:06
  • $\begingroup$ So the residue at z=0 order 2 is -3 using your formula. Did i do it right? $\endgroup$ – bow123 Oct 22 at 17:31
  • $\begingroup$ Yes. That's correct. $\endgroup$ – Mostafa Ayaz Oct 23 at 17:17
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A function "analytic" at a point can be written as "Taylor's series" with all positive powers. A point is a "pole of order n" at a point if its power series, about that point, has negative powers down to -n. And the "residue" at that point is the coefficient of the -1 power.

Here, $f(z)= \frac{1}{z^2(z-1)^3}$ clearly has pole of order 2 at z= 0 and a pole of order 3 at z= 1.

To write f(z) as a Laurent series about z= 0, write it as $f(z)= \frac{1}{z^2}\left(\frac{1}{(z-1)^3}\right)$. $\frac{1}{(z-1)^3}$ is analytic at z= 0 so can be written as a Taylor's series about z= 0. But multiplying that by $\frac{1}{z^2}$ will give a Laurent series where the highest power of z is -2. There is no term having power -1 so the residue at z= 0 is 0. Similarly for the residue at z= 1.

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  • $\begingroup$ Isn't the residue at z=0 with order 2 = -3? I used the formula given my mostafa below $\endgroup$ – bow123 Oct 22 at 17:29
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Another way, perhaps you learned in Calculus, is "partial fractions".

$$ \frac{1}{z^2(z-1)^3} = \frac{1}{(z-1)^3}+\frac{-2}{( z-1 ) ^{2}}+\frac{\color{red}{3}}{( z-1)} +\frac{-1}{z^2}+\frac{\color{blue}{-3}}{z} $$ and we read the residues from there. The residue at $z=1$ is $\color{red}{3}$ and the residue at $z=0$ is $\color{blue}{-3}$

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