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Question Number 213604 by issac last updated on 10/Nov/24

show that \int_{\boldsymbol C} e^{z^{3}} d z = 0

where \boldsymbol C is any simple closed contour .

Evaluate the integral

\int_{C_{1}} f (z) d z, \int_{C_{2}} f (z) d z

where f (z) = (y - x) - 3 x^{2} \boldsymbol i

C_{3} = O A; z (y) = x + \boldsymbol i y = \boldsymbol i y, (0 \leq y \leq 1)

C_{1} = A B; z (x) = x + \boldsymbol i y = x + \boldsymbol i, (0 \leq x \leq 1)

C_{2} = O B; z (x) = x + \boldsymbol i y = x + \boldsymbol i x, (0 \leq x \leq 1)

{Let}^{'} s C be the quadrant

z = 2 e^{\boldsymbol i θ}, 0 \leq θ \leq \frac{π}{2}

show that

∣ \int_{C} \frac{z^{} + 4}{z^{3} - 1} d z ∣\leq \frac{6 π}{7}

Let C be any simple closed contour

described in the positive sense in the z plane

and write

g (z) = \int_{C} \frac{s^{3} + 2 s}{{(z - 2 s)}^{3}} d s

show that g (z) = 6 π \boldsymbol i z when z is inside C

show that g (z) = 0 when z is outside C