written 2.2 years ago by |
Given,
$\oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz $ ; where C is |z| = 1.
Now,
Poles $z-\frac{\pi}{6} =0$
$z=\frac{\pi}{6}$
The point $\frac{\pi}{6}$ lies inside the circle c.
Hence by Cauchy's Integral Formula, we have.
$\implies \oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz = ?$
$\implies \frac{2\pi i}{2!} f^{2} (z) $
$\implies \frac{2\pi i}{2} f''(z) $
$\implies \oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz\ = \pi if''(z) $
Now, Since -
$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} $
$f''(z=\frac{\pi}{6})=30 sin^4(\frac{\pi}{6})cos^2(\frac{\pi}{6})-6 sin ^6 (\frac{\pi}{6})$
$f''(z=\frac{\pi}{6})=30(\frac{1}{16}) (\frac{3}{4})-6(\frac{1}{64})$
$f''(z=\frac{\pi}{6})=\frac{45}{32}-\frac{3}{32}$
$f''(z=\frac{\pi}{6})=\frac{21}{16}$
$\therefore\ \oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz =i\pi \frac{21}{16} = \frac{21}{16} i \pi Ans.$