### Author Topic: TUT0401 QUIZ3  (Read 1027 times)

#### Weiyin Wu

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##### TUT0401 QUIZ3
« on: February 06, 2020, 01:57:50 PM »
Let $\gamma_{1}$ be the semicircle from 1 to -1 through i and $\gamma_{2}$ the semicircle from 1 to -1 through -i
Compute $\int_{\gamma_{1}} z^2 dz$ and  $\int_{\gamma_{2}} z^2 dz$
Can you account for the fact that they are equal?
$$r_{1}(t) = e^{it} (0\leq t \leq\pi)$$
$$r'_{1}(t) = ie^{it}$$
$$\int_{\gamma_{1}} z^2 dz = \int_{0}^{\pi} (e^{it})^{2}\cdot ie^{it}dt$$
$$=i\int_{0}^{\pi} e^{3it}dt$$
$$=\frac{1}{3}(e^{3i\pi}-e^{0})$$
$$=\frac{1}{3}(\cos (3\pi) +i\sin (3\pi)-1)$$
$$=\frac{-2}{3}$$
Similarly,
$$r_{2}(t) = e^{it} (-\pi\leq t \leq 0)$$
$$r'_{2}(t) = ie^{it}$$
$$\int_{\gamma_{2}} z^2 dz = \int_{0}^{-\pi} (e^{it})^{2}\cdot ie^{it}dt$$
$$=\frac{1}{3}(e^{-3i\pi}-e^{0})$$
$$=\frac{1}{3}(\cos (-3\pi) +i\sin (-3\pi)-1)$$
$$=\frac{-2}{3}$$
Yes, $\int_{\gamma_{1}} z^2 dz = \int_{\gamma_{2}} z^2 dz$ since $\gamma_{1}$ and $\gamma_{2}$ are in the same direction, $z^{2}$ is analytic and the region (a circle) is close.