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$I=\int_{L} \frac{z}{\bar{z}}dz$, where L is $1\le|z|<2$, $Im(z)\ge 0$ traversed counterclockwise. thanks for helping.

$I=\int_{L} \frac{z}{\bar{z}}dz=\int_{L}{z^2\over |z|}dz$,

Myshkin
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1 Answers1

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Let

$C_1:\,z=2e^{i\theta}~~;~~0\leq\theta\leq\pi$.

$C_2:\,z=t~~;~~-2\leq t\leq-1$.

$C_3:\,z=e^{i\theta}~~;~~0\leq\theta\leq\pi$.

$C_4:\,z=t~~;~~1\leq t\leq2$. \begin{eqnarray} I&=&\int_{L}\frac{z}{\bar{z}}dz\\ &=&\int_{C_1}+\int_{C_2}-\int_{-C_3}+\int_{C_4}\\ &=&\int_0^\pi e^{2i\theta}2ie^{i\theta}d\theta+\int_{-2}^{-1}dt-\int_0^\pi e^{2i\theta}ie^{i\theta}d\theta+\int_{1}^{2}dt\\ &=&\dfrac{-4}{3}+1-\dfrac{-2}{3}+1\\ &=&\dfrac{4}{3} \end{eqnarray}

Nosrati
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  • the answer is $4/3$ – Myshkin Mar 06 '17 at 05:15
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    @MyGlasses Why to make things so hard in $;C_3;$ ? Just parametrize $;z=e^{i\theta};,;;0\le \theta\le\pi;$, evaluate the integral and then change the sign !:$$\int_0^\pi\frac{e^{i\theta}}{e^{-i\theta}}ie^{i\theta}d\theta=\left.\frac13e^{3i\theta}\right|_0^\pi=-\frac23\pi$$ I think this way is easier and, it seems to me, there's a mistake in that integral. – DonAntonio Mar 06 '17 at 08:19
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    @miosaki Sorry. Regard to very useful comment by DonAntonio I edited the solution. – Nosrati Mar 06 '17 at 08:37