# Image of Circular Arc under composition of complex mapping.Help

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.

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 $|z|=1$.
• yes my bad it was a typo it suppose to be $h(z) = \frac14e^{i\pi/4}z$ – user143612 May 8 '14 at 3:21
• The idea of how $h$ works is that if $h(z)=cz$ with $c=ae^{i\phi}$ then $h(re^{i\theta})=are^{i(\theta +\phi)}$ – MPW May 8 '14 at 23:12