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?
Best Answer
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 ...?)