Operations and transformation with two complex numbers

Problem

Let $z_1=-3-2i$ and $z_2=4+6i$. a) Express them in Cartesian, trigonometric and exponential form. b) Compute $\bar z_1 z_2$ and $\dfrac{\bar z_1}{z_2}$. c) Write the complex function $f(z)$ that translates $3$ downward, then $3$ to the right, then rotates by $\dfrac{\pi}{3}$ clockwise.

Solution

a) $|z_1|=\sqrt{13}$, $\arg z_1\approx213{,}7^\circ$: $z_1=\sqrt{13}\,(\cos213{,}7^\circ+i\sin213{,}7^\circ)=\sqrt{13}\,e^{i\,3{,}729}$. $|z_2|=\sqrt{52}=2\sqrt{13}$, $\arg z_2\approx56{,}31^\circ$: $z_2=2\sqrt{13}\,e^{i\,0{,}983}$. b) $\bar z_1=-3+2i$, so $\bar z_1 z_2=(-3+2i)(4+6i)=-24-10i$; $\;\dfrac{\bar z_1}{z_2}=\dfrac{(-3+2i)(4-6i)}{16+36}=\dfrac{26i}{52}=\dfrac12 i$. c) Downward translation: $z-3i$; to the right: $z+3-3i$; clockwise rotation by $\tfrac{\pi}{3}$: multiplication by $e^{-i\pi/3}$: $f(z)=e^{-i\pi/3}\,(z+3-3i).$ $\boxed{\,\bar z_1 z_2=-24-10i,\quad \dfrac{\bar z_1}{z_2}=\tfrac12 i,\quad f(z)=e^{-i\pi/3}(z+3-3i)\,}$