I need to calculate this integral: $$\oint _{\left | {5} \right |}{\frac{dz}{z-\left (\frac{1}{z} \right )^{2006}}}$$
I got the following hint:
first prove that: $$\forall R > 1\>\>\>\oint _{\left | {z} \right |=5}{\frac{dz}{z-\left (\frac{1}{z} \right )^{2006}}}=\oint _{\left | {z} \right |=R}{\frac{dz}{z-\left (\frac{1}{z} \right )^{2006}}}$$ then prove that: $$\left | \oint _{\left | {z} \right |=R}{\frac{dz}{z-\left (\frac{1}{z} \right )^{2006}}}-\oint _{\left | {z} \right |=R}{\frac{1}{z}}\right | \underset{R\rightarrow \infty}{\rightarrow}0$$
The first statement quickly follows from the Cauchy’s theorem, because the integrated function is holomorphic in the region $R > 1$.
For the second part use $$ \frac 1{z - z^{-2006}} - \frac 1z = \frac {z^{2006}}{z^{2007} - 1} - \frac 1z = \frac {z^{2007} - z^{2007} + 1}{z(z^{2007} - 1)} = \frac 1{z(z^{2007}-1)} $$ to prove for $R > 2$ $$ \left|\oint_{|z| = R} \left[\frac{1}{z - z^{-2006}} - \frac 1z\right] \, dz \right| \leq 4\pi R^{-2007}. $$ It’s a rough estimate, but it’s enough.