Evaluation of line integrals not using winding number 
These are exercises from Conway. I calculated them with the notion of winding number and it was very simple. The value of integrals are $\pi$$i$ and $0$ each. However I am curious about how to directly compute the integrals. I think I need to do some integration by substitution, but can't find a way. Could anyone help me to directly compute these integrals?
 A: You can directly compute
$$\oint_{|z|=1}\frac{dz}{z-t}$$
for all $t$ with $|t|\neq 1$; this of course gives the same answer as the residue theorem, and can be used to evaluate your integrals by translation and scaling and partial fractions.
For $|t|>1$, the Taylor series $\frac{1}{z-t}=-\sum_{k\geq 0}z^kt^{-k-1}$ can be integrated term by term, giving zero.
For $|t|<1$, the Laurent series $\frac{1}{z-t}=z^{-1}\sum_{k\leq 0}z^kt^{-k}$ can be integrated term by term, giving $2\pi i$ from the $z^{-1}$ term.
A: Hint:
One (long) way is using Green's theorem with writing
$$\int_\gamma\dfrac{1}{z^2-1}dz=\int_\gamma\dfrac{x^2-y^2-1-2ixy}{(x^2-y^2-1)^2+(2xy)^2}(dx+idy)=\int_\gamma P_1dx+Q_1dy+i\int_\gamma P_2dx+Q_2dy$$
where $R$ is the region interior of $\gamma$.
A: Hint (not completely sure how to handle the branch cut problem which was raised by MyGlasses, but this leads to the correct result for the first integral): Evaluate the line integral for $z(t)=1+\exp(it)$.
$$\int_{\gamma(t)}\frac{1}{z^2-1}\dfrac{dz}{dt}dt=\int_{0}^{2\pi}\frac{1}{(1+\exp(it))^2-1}i\exp(it)dt$$
$$=\int_{0}^{2\pi}\frac{1}{1+2\exp(it)+\exp(2it)-1}i\exp(it)dt=\int_{0}^{2\pi}\frac{idt}{2+\exp(it)}$$
Use the substitution $u=\exp(it) \implies du =i\exp(it)dt \implies idt = \dfrac{du}{u}$ and use pratial fractions. 
