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Let $\gamma$ be the half unit circle joining $1+i$ and $1-i$ clockwise. By direct parametrization compute,

$\int_{\gamma}$ $\sqrt{z-1}$ $dz$

where the principal branch of the square root is used.

So I first parametrized $\gamma$ as follows,

$\gamma(t)$ = $(1-i)+2it$, $0\leq t \leq 1$
            $1-\exp(\frac{-\pi t i}{2})$, $1\leq t \leq 3$

Now I will have two integrals over two different paths,

$\int_{\gamma}$ $\sqrt{z-1}$ $dz$ = $\int_{\gamma_{1}}$$\sqrt{z-1}$+$\int_{\gamma_{2}}$$\sqrt{z-1}$

Now I know the principal branch of the square root of $z-1$ is defined as

$\exp(\frac{1}{2}\log(z-1))$

where log is a branch of the logarithm. Here I will use the logarithm function as defined in the slit complex plane, writing $z=r\exp(i\theta)$, by

$\text{Log}(z) = \log(r) + i\theta$

where log is the usual logarithm over the real numbers and $-\pi < \theta < \pi$. Am I on the right track for this problem?

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  • $\begingroup$ $\gamma$ as defined by that formula doesn't look like a half unit circle at all; it looks like the union of two line segments. $\endgroup$ – user14972 Oct 13 '15 at 3:55
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Let's parameterize the curve $\gamma$ as $z=1+e^{it}$, where $t$ starts at $\pi/2$ and ends at $-\pi/2$. Then,

$$\begin{align} \int_{\gamma}\sqrt{z-1}\,dz&=\int_{\pi/2}^{-\pi/2}\sqrt{e^{it}}ie^{it}\,dt\\\\ &=\frac23 \left.\left(e^{it}\right)^{3/2}\right|_{\pi/2}^{-\pi/2}\\\\ &=-i\frac{2\sqrt{2}}{3} \end{align}$$

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