Given:
Calculate $\int_{|z|=2}\sqrt{z^2-1} dz$
Hint:$\sqrt{z^2-1}=z\sqrt{1-\frac{1}{z^2}}:=z\exp(\frac{1}{2}\log(1-\frac{1}{z^2}))$
I tried it for several hours and didn't manage to get anywhere i would be happy to get another hint for solving this.
Edit:
I thought maybe using the formula $(1+z)^k=\sum_{0}^{\infty}\frac{k(k-1)...(k-n+1)}{n!}z^n$
and i got to this after some calcualtions
$\int_{|z|=2}\frac{z}{2}+\frac{1}{4z}+o(z^{-3})dz$
Your hint leads to the Laurent expansion $$\eqalign{\sqrt{z^2-1}&=z\left(1-{1\over z^2}\right)^{1/2}=z\left(1-{1\over2z^2}-{1\over8z^4}-{1\over16z^6}-\ldots\right)\cr &=z-{1\over2z}-{1\over8z^3}-{1\over16z^5}-\ldots\qquad\bigl(|z|>1\bigr)\ ,\cr} $$ which converges uniformly, e.g., on an annulus ${3\over2}\leq|z|\leq3$. Here I have developed $\left(1-{1\over z^2}\right)^{1/2}$ into a binomial series, using the standard formula $$(1+t)^\alpha=\sum_{k=0}^\infty{\alpha\choose k} t^k\qquad\bigl(|t|<1\bigr)\ .$$It is now easy to determine the value of the desired integral.
