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I've been having a little trouble here figuring this out. How would one sketch the image of the sector $0 < \theta < \pi/6$, where $\theta$ is the polar angle of the complex number $z$, under the complex power function $w=z^a$ when $a=\frac{3}{2}$? Also, what would happen if $a=i$? It seems the only thing I've been able to get around so far is the case where $a$ is an integer. I would appreciate some guidance here.

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Replace the $z$ with $r\exp(i\theta)$, separate out real and imaginary parts, and you now have parametric equations for a family of curves... –  J. M. Sep 11 '11 at 1:10
    
Putting under the rug (as some interventions here seem to do) the key fact that there is no way to define continuously $z^{3/2}$ on the whole complex plane leaves me uneasy. We all know that $r\exp(i\theta)=r\exp(i(\theta+2\pi))$ and that applying the seemingly obvious operation $z\to z^{3/2}$ to the LHS and to the RHS does not give the same result, don't we? The same objection (only possibly worse) applies to $z^i$. –  Did Sep 11 '11 at 9:37
    
@Didier: I guess we could make a branch cut at the ray that is $\pi/6$ radians in angle measure from the positive real axis to make the function continuous? –  Libertron Sep 11 '11 at 16:02

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Proceeding from @J. M.'s comment: For $a=3/2$ $w=r^{3/2} e^{\frac{3}{2}i \theta}$ or $w=r^{3/2} \left( {1+ \frac{3i \theta}{2} -\frac{9 {\theta}^2}{8}-\frac{27 i {\theta}^3}{48} + \frac{81 {\theta}^4}{384}+\ldots} \right)$ observe the curve plot at Wolfram Alpha

For $a=i$: $w=r^i e^{-\theta}$ See the Curve

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For which $\theta$? See my comment to the question. –  Did Sep 11 '11 at 9:38

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