written 4.9 years ago by | • modified 4.9 years ago |
SOlution:
$\oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz$
Poles $z-\frac{\pi}{6} =0$
$z=\frac{\pi}{6}$ and $|z| =1$ is a circle ,with center at origin and radius 1.The point $\frac{\pi}{6}$ lies inside the circle c.
Hence by cauchy formula .
$\int \frac {f(z)}{(z-z_0)}=\frac {2\pi i }{(n-1)!}f^{n-1}(z_0)$
$\int \frac {sin^6z}{(z-\frac{\pi }{6})^3}=\frac {2\pi i }{(3-1)!}f^{3-1}(z_0)$
Here n=3
=$ \frac{2\pi i}{2!} f^{2} (z_0) $
=$ \frac{2\pi i}{2} f''(z_0) $
$\oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz$ =$\pi if''(z_0) $
$f(z) = sin^6 z $
$f'(z)= 6sin^5z.cos z $
$f''(z) = 6.5 sin^4z .cosz.cosz+6sin^5z(-sin z)$
=$30sin^4z .cos ^2z-6sin^6z$
Put $z= \frac{\pi}{6} $
$f''(z=\frac{\pi}{6})=30 sin^4(\frac{\pi}{6})cos^2(\frac{\pi}{6})-6 sin ^6 (\frac{\pi}{6})$
$f''(z=\frac{\pi}{6})=30(\frac{1}{16}) (\frac{3}{4})-6(\frac{1}{64})$
$f''(z=\frac{\pi}{6})=\frac{45}{32}-\frac{3}{32}$
$f''(z=\frac{\pi}{6})=\frac{21}{16}$-----(3)
Put (3) in eqn(2)
$\oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz =i\pi \frac{21}{16} = \frac{21}{16} i \pi$