1. ## Complex Mapping

Find the image of the following region

under the action of $f(z)=z^5$.

I am horrible with mappings. I wrote $z^5=r^5 \cdot e^{5i\theta_0}$, and tried points in both regions, and then I ended up with the whole complex plane as the output and I don't think this is right. How do I find the image of this region? Thanks for the help.

2. The line y= 3x corresponds to the set of complex number a+ 3ai= a(1+ 3i) for a any real number (do you see why?). $z^5= (a+ 3ai)^5= a^5(1+ 3i)^5$. Now you could, if you wanted, multiply that out directly or use the "polar form". It will, in any case, be a constant, say u+ vi, so $z^5= a^5(u+ vi)$ which gives "parametric" equations [/tex]y= va^5[/tex] and $x= ua^5$. $a^5= \frac{x}{u}$ (as long as u is not 0) so $y= v\frac{x}{u}= \frac{v}{u} x$ another straight line. Similarly for y= x/2 which corresponds to the set of complex numbers $a+ \frac{1}{2}ai$.

That is, $z^5$ maps that region, bounded by straight lines, into another region bounded by straight lines. You will need to compute $(1+ 3i)^5$ and $(1+ \frac{1}{2})^5$ to determine exactly which straight lines.