I'm stumbled upon a question when computing Residue.
I want to compute the integral $\int_{\gamma} \frac{\cos(z)}{z-1} dz$ with help of residue.
My solution is rather short since I can see directly that this has a pole of order 1 which gives me the residue $\cos(1)$ using the formula $$
\mathrm{Res}_{z_0} f(z) = \frac{1}{(n-1)!} \lim_{z\to z_0} \frac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}} (z-z_0)^nf(z)
$$
However, in other exercises such as computing the integral $\int_{\gamma} \frac{\cos(z)}{z} dz$, I solved this with finding the coefficient $c_{-1}$, which I have problem with doing on the first integral since I don't see how the first factor $(z-1)$ would disappear in the Laurent series expansion.
Would anyone mind helping me out on how to do that? Or give me an hint?
Best Answer
You have\begin{align}\frac{\cos(z)}{z-1}&=\frac{\cos\bigl((z-1)+1\bigr)}{z-1}\\&=\frac{\cos(z-1)\cos(1)-\sin(z-1)\sin(1)}{z-1}\\&=\cos(1)\frac{\cos(z-1)}{z-1}-\sin(1)\frac{\sin(z-1)}{z-1}\\&=\cos(1)\left(\frac1{z-1}-\frac12(z-1)+\frac1{4!}(z-1)^3+\cdots\right)+\\&{}\qquad-\sin(1)\left(1-\frac1{3!}(z-1)^2+\frac1{5!}(z-1)^4-\cdots\right)\\&=\frac{\cos(1)}{z-1}-\sin(1)-\frac{\cos(1)}2(z-1)+\frac{\sin(1)}{3!}(z-1)^2+\cdots\end{align}