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Using Cauchy's Integral formula evaluate $\int_{c}\frac{sin^6z}{(z-\pi/6)^3}dz$ where c is |z|=1
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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} $

$f''(z=\frac{\pi}{6})=30 sin^4(\frac{\pi}{6})cos^2(\frac{\pi}{6})-6 sin ^6 (\frac{\pi}{6})$

$f''(z=\frac{\pi}{6})=30(\frac{1}{16}) (\frac{3}{4})-6(\frac{1}{64})$

$f''(z=\frac{\pi}{6})=\frac{45}{32}-\frac{3}{32}$

$f''(z=\frac{\pi}{6})=\frac{21}{16}$


$\therefore\ \oint_c \frac{sin^6z}{(z-\frac{\pi}{6})^3} dz =i\pi \frac{21}{16} = \frac{21}{16} i \pi Ans.$

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