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$\oint \frac{e^{2z}}{(Z+1)^4}dz$ where c is the circle |z-1|=3.

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14kviews

$\oint \frac{e^{2z}}{(Z+1)^4}dz$ where c is the circle |z-1|=3.

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written 6.4 years ago by | • modified 6.1 years ago |

The path of integration is the circle with centre (1, 0) & radius =3

The points at which the function is not analytic are z= -1 which lies inside C .

By Cauchy’s Integral formula, value of the integral is given by

$∮ \frac{e^{2z}}{(z+1)^4} \ dz$

= $ \frac{2πi}{3!} [\frac{d^3}{dz^3} e^{2z} ]_{z=(-1)}$

= $ \frac{4πi}{3!} [\frac{d^2}{dz^2} e^{2z} ]_{z=(-1)}$

= $ \frac{8πi}{3!} [\frac{d}{dz} e^{2z} ]_{z=(-1)} = \frac{4πi}{3} [2 e^{2z} ]_{z=(-1)}$

= $ \frac{8πi}{3} e^{-2}$

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