# Laurent Series, Taylor Series, and Order of Poles. A tale of confusion.

For $\int_{C}\frac{\sin(z)}{(z^2 + 2z - 3)^2} dz$, where $C = \{|z|=2\}$, we have singularities are $z = -3$, $z = 1$. So only $z = 1$ is contained within the contour. This singularity has order $m=2$ right? The $\sin(1) \neq 0$ so that means we don't need to apply the taylor series for the numerator. The only way for the function to continue being analytic is by multiplying $f(z)$ by $(z-1)^2$.

Is this correct?

I'm still not very sure of $\textit{why}$ when the numerator is $0$ at the singularity point, that means we can simplify the function via Taylor series / Laurent Series.

For example,

$$\frac{e^z - 1}{z^6}$$ has a singularity at $z=0$. Initially, I thought this pole would have order $6$, but it is actually order $5$. Note that at $z=0$, the numerator is $0$. This hints at simplifying via TS/LS. Why?

The Taylor Series of $e^z - 1$ is: $z + \frac{z^2}{2} + \frac{z^3}{3!}+\cdots$ Since we define the residue of a function to be the $a^{-1}$ coefficient term of the laurent series of the function, we look for for what will give a $z^{-1}$ term after dividing by $z^6$ in the expansion of $e^z -1$. This of course is $z^5$ which leads us to find that $\frac{1}{5!}$ is the corresponding residue.

Now, why is this all the Laurent Series of our function? With non-negative powers, are the Laurent and Taylor series the same?

To evaluate the integral, I'd proceed this way. The integrand equals

$$\frac{\sin z}{(z+3)^2}\cdot \frac{1}{(z-1)^2}.$$

We know that first quotient, call it $f,$ is analytic at $1.$ Hence it will have a Taylor series there that looks like $f(z) = f(1) +f'(1)(z-1) + \cdots.$ Thus our integrand, near $1,$ equals

$$\frac{f(1)}{(z-1)^2} + \frac{f'(1)}{z-1} + \cdots.$$

It follows that the residue of the integrand is $f'(1),$ and the integral equals $2\pi i\cdot f'(1).$ Calculate $f'(1)$ and you're done.

(I don't understand some of the questions you raise. Perhaps if you asked a specific question about what I did ...?)

• Okay, here is a more direct question, I understand how to find the Residue. I am confused about what the order of the singularity is. For example:$\frac{\sin^2(z)}{z^3(z-1)}$, $Res(f,0) = -1$ since $\sin^2(z) = z^2 - \frac{z^4}{3} + \cdots$ and $\frac{z^2}{z^3(z-1)} = \frac{z^{-1}}{-1}$ at $z=0$. However, I don't understand what the order of the singularity, $z=0$, should be. Commented May 1, 2015 at 9:04
• $f$ has a pole at $0$ if, near $0,$ $$f(z) = \frac{a_N}{z^N}+ \cdots \frac{a_1}{z}+b_0 + b_1z + \cdots ,$$ where $N\ge 1$ and $a_N\ne 0.$ In this case we say the order of the pole is $N.$ We don't need to know the Laurent expansion in order to know the order. Fact: $f$ has a pole or order $N$ at $0$ iff $\lim_{z\to 0} z^Nf(z)$ exists and is a finite nonzero complex number. So in your example, it's pretty clear that the order of the pole is $1.$ Just multiply by $z$ and you get $-1$ for the limit.
– zhw.
Commented May 1, 2015 at 17:51

Concerning your questions about order of a singularity (pole): A function $f$ which is analytic in a punctured neighborhood $\dot U$ of $0$ has a pole of order $r\geq1$ at $0$ if there is an analytic function $g:\>U\mapsto{\mathbb C}$ with $g(0)\ne0$ such that $$f(z)={g(z)\over z^r}\qquad(z\in\dot U)\ .\tag{1}$$ The values $r$ and $g(0)$ then tell how fast $f$ goes to $\infty$ when $z\to0$.

The above has a priori nothing to do with the residue of $f$ at $0$, but it plays a rôle when it comes to computing this residue not via an integral, but by algebraic means. The function $f$ in $(1)$ has a Laurent expansion at $0$ of the form $$f(z)={a_r\over z^r}+{a_{r-1}\over z^{r-1}}+\ldots+{a_2\over z^2}+{a_1\over z}+\sum_{k=0}^\infty c_k z^k\qquad(z\in\dot U)\ .$$ It so happens that the residue in question is $=a_1$, and that it can be painful to access this value when $r$ is large. Therefore it pays to know exactly the order of the pole before entering the computations.