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How do we proceed for the following complex line integral?

$$\int\limits_\gamma |z|\:dz$$

where $\gamma$ is the half circular $|z|=1$, $0\leq \arg (z) \leq \pi$

taking $z=1$ as the initial point.

Any hints would be helpful. Thanks.

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Substitute $z = e^{i\theta}$. The limits become $0$ to $\pi$

$|z| = 1$ and $dz$ = $ie^{i\theta}\:d\theta$

$$\int\limits_\gamma |z| \:dz = \int\limits_0^\pi ie^{i\theta} \:d\theta$$

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  • $\begingroup$ Thank you.. one more question... that if the angle were $\frac{-\pi}{2} \leq \arg z \leq \frac{\pi}{2}$ and z = - i, was the initial point? we just need to change the limit of the integration by $\frac{- \pi}{2}$ to $\frac{\pi}{2}$?? $\endgroup$ – rndflas Dec 16 '15 at 10:33
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    $\begingroup$ yes, you are right $\endgroup$ – glip-glop Dec 16 '15 at 10:38

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