First Principles AI

Solution

Given:

z_1=9\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right),\quad z_2=5\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)

Convert to rectangular form when adding/subtracting, and use polar rules for multiplication/division.

\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2},\ \sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}

z_1=9\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)=-\frac{9\sqrt{2}}{2}-\frac{9\sqrt{2}}{2}i

\cos\frac{\pi}{3}=\frac{1}{2},\ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}

z_2=5\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=\frac{5}{2}+\frac{5\sqrt{3}}{2}i

\boxed{z_1+z_2=\frac{5-9\sqrt{2}}{2}+\frac{5\sqrt{3}-9\sqrt{2}}{2}i}

\boxed{z_1-z_2=-\frac{9\sqrt{2}+5}{2}-\frac{9\sqrt{2}+5\sqrt{3}}{2}i}

z_1z_2=(9\cdot 5)\left(\cos\left(\frac{5\pi}{4}+\frac{\pi}{3}\right)+i\sin\left(\frac{5\pi}{4}+\frac{\pi}{3}\right)\right)

\frac{5\pi}{4}+\frac{\pi}{3}=\frac{15\pi}{12}+\frac{4\pi}{12}=\frac{19\pi}{12}

\cos\frac{19\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4},\quad \sin\frac{19\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4}

\boxed{z_1z_2=\frac{45(\sqrt{6}-\sqrt{2})}{4}-\frac{45(\sqrt{6}+\sqrt{2})}{4}i}

\frac{z_1}{z_2}=\frac{9}{5}\left(\cos\left(\frac{5\pi}{4}-\frac{\pi}{3}\right)+i\sin\left(\frac{5\pi}{4}-\frac{\pi}{3}\right)\right)

\frac{5\pi}{4}-\frac{\pi}{3}=\frac{15\pi}{12}-\frac{4\pi}{12}=\frac{11\pi}{12}

\cos\frac{11\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4},\quad \sin\frac{11\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}

\boxed{\frac{z_1}{z_2}=-\frac{9(\sqrt{6}+\sqrt{2})}{20}+\frac{9(\sqrt{6}-\sqrt{2})}{20}i}