Given two complex numbers $z=r\exp(i\theta),z'=r'\exp(i\theta')$, I would like to prove that $zz'$ is constructible using straight-edge and compass. I am stuck on proving the constructability of a ray at an angle $\theta + \theta'$. I can't show my attempts because they are figures. Any assistance?
Update
I think I figured it out (in the case $\theta, \theta' > 0$, but the other cases are similar). Draw the circle of center $0$ and radius $r$ and let $w$ be the intersection of this with the positive $x$-axis. This circle intersects the ray $0z'$ in $z''$. Now draw the circle of center $z''$ and radius $|z' -w|$. This intersects the first circle at $z'''$. The required ray is $0z'''$.
Let me give a construction which works as stated in the case where $\theta+\theta'$ is no larger than $\pi$; adjustments will be neeced otherwise.
Suppose you have two segments $BA,BC$ meeting at angle $\theta$, and two segments $B'A',B'C'$ meeting at angle $\theta'$.
Construct a point $A''$ on the ray $\overrightarrow{BA}$ so that $BA''$ is congruent to $B'A'$.
Construct a triangle $BA''C''$ which is congruent to $B'A'C'$. There are two choices for this triangle; the angle between the segments $BC''$ and $BC'$ will equal either $\theta + \theta'$ or $|\theta - \theta'|$. You can constructily describe the difference between these two, and choose the correct one to produce $\theta+\theta'$.