# Find the Zeroes, poles and their orders and the residue at each pole?

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

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?

• $z=0$ and $z=1$ are not zeroes, they are poles. – GEdgar Oct 22 at 17:01
• sorry i meant poles – bow123 Oct 22 at 17:09

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.

• Can you show how this would work for either z=0 or z=1? – bow123 Oct 22 at 17:02
• Sure. I added some more details. – Mostafa Ayaz Oct 22 at 17:06
• So the residue at z=0 order 2 is -3 using your formula. Did i do it right? – bow123 Oct 22 at 17:31
• Yes. That's correct. – Mostafa Ayaz Oct 23 at 17:17

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.

• Isn't the residue at z=0 with order 2 = -3? I used the formula given my mostafa below – bow123 Oct 22 at 17:29

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 $$\color{blue}{-3}$$