Finding residue of complex function, the result is different when using Laurent series and residue theorem.

Find the residue of $$f(z)=\dfrac{z^3+2z+1}{(z-1)(z+3)}$$ on simple pole $$z=1$$.

If I using residue theorem, I have \begin{align} \underset{z=1} {\operatorname{Res}} f(z) = \lim\limits_{z\to 1} (z-1)\dfrac{z^3+2z+1}{(z-1)(z+3)}=\dfrac{1+2+1}{4}=1. \end{align}

If I using Laurent series method, I have \begin{align} f(z)&=\dfrac{z^3+2z+1}{(z-1)(z+3)}\\ &=(z-2)+\dfrac{9z-5}{(z-1)(z+3)}\\ &= -1+(z-1)+\dfrac{9(z+3)-32}{(z-1)(z+3)}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z+3}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{z-1}\cdot\dfrac{1}{z-1+4}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\cdot\dfrac{1}{1+\dfrac{4}{z-1}}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-1)^n \left(\dfrac{4}{z-1}\right)^n\\ &= -1+(z-1)+\dfrac{9}{(z-1)}-\dfrac{32}{(z-1)^2}\sum\limits_{n=0}^{\infty} (-4)^n \left(z-1\right)^{-n}\\ &= -1+(z-1)+\dfrac{9}{(z-1)}+\sum\limits_{n=0}^{\infty} (-32)(-4)^n \left(z-1\right)^{-n-2}. \end{align} Now I have coefficient of $$(z-1)^{-1}$$ is $$9$$, so we can conclude \begin{align} \underset{z=1} {\operatorname{Res}} f(z) =9. \end{align} My question

When I using residue theorem and Laurent series method, why the result is distinct? What my mistake in my work?

• I think you can't use Taylor expansion of $\frac{1}{1 +\frac{4}{z - 1}}$, because $|\frac{4}{z-1}| \to \infty$ at $1$, but you need $\frac{4}{z-1}$ to be close to zero to use it. – Nik Pronko Apr 28 '20 at 13:26

1 Answer

The "Laurent series method" had an incorrect conclusion going from the fourth to the fifth lines. There was a still an order $$1$$ term on the far right, it just still had to be pulled out.

We can calculate the Laurent series another way. From partial fraction decomposition we have that

$$\frac{4}{(z-1)(z+3)} = \frac{1}{z-1}-\frac{1}{z+3}$$

which means we can rewrite the function as

$$f(z) = \frac{z^3+2z+1}{4(z-1)} - \frac{z^3+2z+1}{4(z+3)}$$

This time the term on the far right has no singularity at $$z=1$$ so it will not contribute to the residue. We will only focus on the term on the left.

Next, shift the variables in the numerator to the desired center:

$$\frac{z^3+2z+1}{4(z-1)} = \frac{(z-1)^3+3(z-1)^2+5(z-1)+4}{4(z-1)}$$

$$= \frac{1}{4}(z-1)^2+\frac{3}{4}(z-1) + \frac{5}{4} + \frac{1}{z-1}$$

which has a residue of $$1$$.