MAT334--2020S > Quiz 3

TUT0401 QUIZ3

(1/1)

**Weiyin Wu**:

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.

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