Answer
$z_1z_2=14+2i$
$\frac{z_1}{z_2}=-\frac{1}{4}+\frac{7}{4}i$
$\frac{1}{z_1}=\frac{3}{25}-\frac{4}{25}i$
Work Step by Step
Step 1. Covert the complex numbers into polar forms: $z_1=3+4i=r_1(cos\theta_1+i\cdot sin\theta_1), r_1=\sqrt {4^2+3^2}=5, tan\theta_1=\frac{4}{3}, sin\theta_1=\frac{4}{5}, cos\theta_1=\frac{3}{5}$. Similarly, $z_2=2-2i=r_2(cos\theta_2+i\cdot sin\theta_2), r_2=\sqrt {2^2+2^2}=2\sqrt 2, tan\theta_2=-1, sin\theta_2=-\frac{\sqrt 2}{2}, cos\theta_2=\frac{\sqrt 2}{2}$
Step 2. Calculate $z_1z_2=r_1r_2(cos(\theta_1+\theta2)+i\cdot sin(\theta_1+\theta_2))$, since $cos(\theta_1+\theta2)=\frac{3}{5}\times \frac{\sqrt 2}{2}+\frac{4}{5}\times \frac{\sqrt 2}{2}= \frac{7\sqrt 2}{10}$ and $sin(\theta_1+\theta2)=\frac{4}{5}\times \frac{\sqrt 2}{2}-\frac{3}{5}\times \frac{\sqrt 2}{2}= \frac{\sqrt 2}{10}$, we have $z_1z_2=10\sqrt 2( \frac{7\sqrt 2}{10}+i\cdot \frac{\sqrt 2}{10})=14+2i$
Step 3. Calculate $\frac{z_1}{z_2}=\frac{r_1}{r_2}(cos(\theta_1-\theta2)+i\cdot sin(\theta_1-\theta_2))$, since $cos(\theta_1-\theta2)=\frac{3}{5}\times \frac{\sqrt 2}{2}-\frac{4}{5}\times \frac{\sqrt 2}{2}= -\frac{\sqrt 2}{10}$ and $sin(\theta_1-\theta2)=\frac{4}{5}\times \frac{\sqrt 2}{2}+\frac{3}{5}\times \frac{\sqrt 2}{2}= \frac{7\sqrt 2}{10}$, we have $\frac{z_1}{z_2}=\frac{5\sqrt 2}{4}( -\frac{\sqrt 2}{10}+i\cdot \frac{7\sqrt 2}{10})=-\frac{1}{4}+\frac{7}{4}i$
Step 4. Calculate $\frac{1}{z_1}=\frac{1}{3+4i}=\frac{3-4i}{(3+4i)(3-4i)}=\frac{3-4i}{25}=\frac{3}{25}-\frac{4}{25}i$
