Im getting really confused looking at past exam style questions evaluating contour integrals... Can anyone help me in the right direction to solve these..
(i) $\displaystyle\int_\gamma \frac{\sin z}{z^4}dz$ where $\gamma(t)=e^{it}$ and $0\leq t\leq2\pi $
(ii) $\displaystyle\int_\gamma \frac{1}{(z-1)(z+3)}dz$ where $\gamma(t)=2e^{it}$ and $0\leq t\leq 2\pi$
Let $\gamma$ be the unit circle in $\mathbb{C}$ transversed in the anti-clockwise direction.
(iii) $\displaystyle\int_\gamma \frac{\cos^2 z}{z^2}dz$
(iv) $\displaystyle\int_\gamma \frac{1}{(2z+1)(z+3)}dz$
Hints (look for the residue(s) at the pole(s) within the integration contour!):
$$\color{red}{(1)}\;\;\;\frac{\sin z}{z^4}=\frac1{z^4}\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots\right)=\frac1{z^3}-\frac1{6z}+\ldots$$
$$\color{red}{(2)}\;\;\;\frac1{(z+3)(z-1)}=\frac1{4(z-1)}\frac1{1+\frac{z-1}4}=\frac1{4(z-1)}\left(1-\frac{z-1}4+\frac{(z-1)^2}{16}+\ldots\right)=\ldots$$
$$\color{red}{(3)}\;\;\;\frac{\cos^2z}{z^2}=\frac1{z^2}\left(1-\frac{z^2}{2!}+\ldots\right)^2=\frac{1}{z^2}\left(1-z^2+\ldots\right)=\ldots$$
$$\color{red}{(4)}\;\;\;\frac{1}{(2z+1)(z+3)}=\frac12\frac1{z+\frac12}\cdot\frac25\frac1{1+\frac25\left(z+\frac12\right)}=$$
$$=\frac1{5\left(z+\frac12\right)}+\left(1-\frac{2}{5}\left(z+\frac12\right)+\frac4{25}\left(z+\frac12\right)^2-\ldots\right)=\ldots$$