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I am doing some basic problems on Complex Analysis and I've just started with integration, I would like to check if my solutions for the following two integrals are correct:

Calculate

1) $\int_{\gamma} \overline zdz$ for $\gamma:[0,2\pi] \to \mathbb C$ given by $\gamma (t)=e^{it}$

2) $\int_{\gamma}|z|^2zdz$ for the following curve (a piece of the unit disk which belongs to the first quadrant)

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My solution:

1) $\int_{\gamma} \overline zdz=\int_0^{2\pi} e^{it}(e^{it})'dt=\int_0^{2\pi} (\cos(t)+i\sin(t))i(-\sin(t)+i\cos(t))dt$

Reordering the last expression, I get that the integral is

$\int_{\gamma} \overline zdz=\int_0^{2\pi} (\sin^2(t)-\cos^2(t))dt+i\int_0^{2\pi} -2\cos(t)\sin(t)dt=-\cos(t)\sin(t)|_0^{2\pi}+i\cos^2(t)|_0^{2\pi}=0$

2) I've separated the domain of integration in three pieces: $\int_{\gamma}|z|^2zdz=\int_{\gamma_1}|z|^2zdz+\int_{\gamma_2}|z|^2zdz\int_{\gamma_3}|z|^2zdz$ where

$\int_{\gamma_1}|z|^2zdz=\int_0^{\frac{\pi}{2}} (e^{it})^2e^{it}ie^{it}dt$

$\int_{\gamma_2}|z|^2zdz=-\int_0^1 (it)^2(it)idt$

$\int_{\gamma_3}|z|^2zdz=\int_0^1 t^2tdt$

I would like to know if what I've done is correct, specially with the last integral, if I've separated the region of the disk correctly. Thanks in advance.

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    $\begingroup$ In the first integral, you integrated $z$ instead of $\overline{z} = e^{-it}$. In the second, $\lvert z \rvert^2 = z\overline{z}$, not $= z^2$. $\endgroup$ – Daniel Fischer May 6 '14 at 18:03
  • $\begingroup$ Thanks for the correction, the first one was a mistake due to distraction. $\endgroup$ – user148122 May 9 '14 at 2:05
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I know this is old at this point, but in case anyone wanders by with the same or a similar question..

For #1, it'd be easier to just keep it in terms of $e$.

\begin{align} \int_\gamma \overline{z} \, dz &= \int_0^{2\pi}\overline{e^{it}}\, dt\\[0.3cm] &= \int_0^{2\pi} e^{-it} \, dt\\[0.3cm] &= -\frac{1}{i} e^{-it}\bigg|_0^{2\pi}\\[0.3cm] &= -\frac{1}{i} \left(e^{-2\pi i} - e^{0}\right)\\[0.3cm] &= -\frac{1}{i} (1 - 1)\\[0.3cm] &= 0 \end{align}

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