2.I.7BComplex MethodsPart IB, 2003(a) Using the residue theorem, evaluate∫∣z∣=1(z−1z)2ndzz\int_{|z|=1}\left(z-\frac{1}{z}\right)^{2 n} \frac{d z}{z}∫∣z∣=1(z−z1)2nzdz(b) Deduce that∫02πsin2ntdt=π22n−1(2n)!(n!)2\int_{0}^{2 \pi} \sin ^{2 n} t d t=\frac{\pi}{2^{2 n-1}} \frac{(2 n) !}{(n !)^{2}}∫02πsin2ntdt=22n−1π(n!)2(2n)!