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Find the image of the circular arc $|z|=2,\ \ 0 \leq Arg(z) \leq \frac{\pi}{2}$ under the composition $f(z) = (h\circ g)(z)$ where $h(z) = \frac14e^{i\pi/4}z$ and $ g(z) = z^2$

This is what I have done,

I said let $ C = |z|=2, \ \ 0\leq Arg(z) \leq \frac{\pi}{2}$ then $r=2 \implies r^2 = 4$ So, the image of $C$ under $g(z)=z^2$ is the semicirle $C' = |z|=4, \ \ 0 \leq Arg(g) \leq \pi $

Now how do I find the image of $C'$ under the mapping $h(z) = \frac14e^{i\pi/4}z$?

I know that if your mapping is $M(z) = az = a(re^{i\theta})$ where $ a > 0 \ $ then you magnify the modulus your function by a factor of $ \ a$ but since h(z) is not exactly in the form of $a(re^{i\theta})$ I'm having some trouble.Any help would be appreciated.

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This must be a typo, since $h$ is constant. If that's really what's intended, the image is the single point $\frac14e^{i\pi/4}$.

But I suspect it may have been intended that $h(z) = \frac14e^{i\pi/4}z$. This would have the effect of rotating your intermediate arc by $\pi/4$ and shrinking it down to the unit circle. Then the final image would be the semicircle with $\pi/4\leq \operatorname{Arg}(z)\leq5\pi/4$ and $|z|=1$.

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  • $\begingroup$ yes my bad it was a typo it suppose to be $h(z) = \frac14e^{i\pi/4}z$ $\endgroup$ – user143612 May 8 '14 at 3:21
  • $\begingroup$ The idea of how $h$ works is that if $h(z)=cz$ with $c=ae^{i\phi}$ then $h(re^{i\theta})=are^{i(\theta +\phi)}$ $\endgroup$ – MPW May 8 '14 at 23:12
  • $\begingroup$ ok I understand..thanks $\endgroup$ – user143612 May 8 '14 at 23:16

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