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Choose appropriate branches of $\log(z)$ to evaluate $\int_\Gamma z^\frac{1}{2} dz $ for the contour, $\Gamma$, in the right half-plane from $z = -3i$ to $z = 3i$

What I have done so far:

$\\$ $\frac{2z^\frac{3}{2}}{3} \Big\rvert_{-3i}^{3i}$ $= (2+2i)\sqrt{3i}$

However, I'm not sure what the branches of $\log(z)$ have to do with solving the integral? If I substitute $|r|e^{i\theta}$ in for $z$ and then evaluate, I get:

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 9ie^{\frac{3i\theta}{2}} d\theta = \frac{18\sqrt{2}}{5}$

Would this be choosing the appropriate branch cut of $log(z)$? If so, how?

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    They come about from the definition of exponentiation: $$z^{1/2}=\exp(\ln(z)/2)$$ – 2017-02-28
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    So I need to evaluate exp(ln(z)/2), instead of what I did above? It should be the same answer, though. Shouldn't it? – 2017-02-28
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    Yes to the first, no to the second. Note that $$\lim_{\theta\to0^+}(e^{i\theta})^{1/2}=1\ne-1=\lim_{\theta\to2\pi^-}(e^{i\theta})^{1/2}$$Assuming principal branches... – 2017-02-28
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    Correct. [Beyond this though, my skill with complex analysis is probably less than yours, so I won't be of any more help :-( ] – 2017-02-28
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    I guess it means the principal branch of $z^{1/2}$. Can you find a function $f(z)$ holomorphic on $Re(z) \ge 0, z \ne 0$ such that $f'(z) = z^{1/2}$ ? @SimplyBeautifulArt – 2017-02-28
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    I think it is a simple solution. We are only in section 3 of chapter 4 in the book. We haven't talked about holomorphic functions yet. – 2017-02-28
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    Ok then let $f(z) = \frac23 z^{3/2}$ (the principal branch) it is analytic on $(0,3i)$ with $f'(z) = z^{1/2}$ (the principal branch) and so $\int_0^{3i} z^{1/2}dz = \int_0^{3i} f'(z)dz = f(z)|_0^{3i} =\frac23 (3 e^{ i \pi / 2})^{3/2} = \frac23 e^{3 i \pi / 4}3^{3/2}$. Again with the principal branch $f(z)$ is analytic on $(-3i,0)$ with $f'(z) = z^{1/2}$ and so $\int_{-3i}^0 z^{1/2}dz = f(z)|_{-3i}^0 = -\frac23(3e^{- i \pi / 2})^{3/2} = -\frac23 e^{-3 i \pi / 4} 3^{3/2}$ @SimplyBeautifulArt – 2017-02-28
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    I'm not sure how a branch of log(z) is used in your solution. – 2017-02-28

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This is the answer I came up with. Please let me know if there are any errors in my logic.

$z^\frac{1}{2} = e^{\frac{1}{2}\log(z)}$ and $z = 3e^{i\theta} $ $\Rightarrow$ $\int_\Gamma z^\frac{1}{2} = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} e^{\frac{1}{2}\log(3e^{i\theta})}(3ie^{i\theta})d\theta$

$=$ $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (3^\frac{3}{2})ie^{\frac{3i\theta}{2}}d\theta = (3^\frac{3}{2})i(\frac{2i}{5}e^{\frac{5i\theta}{2}}) |_\frac{-\pi}{2}^{\frac{\pi}{2}}$

$= \frac{6i\sqrt{6}}{5}$