Question

# Evaluate $\int_{\gamma} f(z) d z.$ $f(z)=z^{2}-i z, \gamma$ is the quarter circle about the origin from 2 to 2i.

Solution

Verified
Step 1
1 of 2

Let

$f(z) = z^2-iz$

Note that $\gamma$ is given by the following parametrization:

$\gamma(t) = 2e^{it} \quad \text{ for } 0 \le t \le \pi/2$

The formula for the integral is

$$$\int_{\gamma}f(z)\dd{z} = \int_a^b f(\gamma(t))\gamma'(t)\dd{t}$$$

Now we have

\begin{align*} & f(\gamma(t)) = (2e^{it})^2-2ie^{it} = 4e^{2it} -2ie^{it}\\ & \gamma'(t) = 2ie^{it} \end{align*}

Substituting into formula (1), we get

\begin{align*} \int_{\gamma}f(z)\dd{z} &= \int_0^{\pi/2}\qty(4e^{2it}-2ie^{it})2ie^{it}\dd{t}\\ &=\int_0^{\pi/2}\qty(8ie^{3it}+4e^{2it})\dd{t}\\ &=\eval{\qty(\dfrac{8}{3}e^{3it}+\dfrac{2}{i}e^{2it})}_0^{\pi/2}\\ &=\eval{\qty(\dfrac{8}{3}e^{3it}-2ie^{2it})}_0^{\pi/2}\\ &=-\dfrac{8}{3}i+2i-\dfrac{8}{3}+2i\\ &=-\dfrac{8}{3}+\dfrac{4}{3}i \end{align*}

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