198. Perform the division:
a) $\frac{2+3 i}{5-7 i}$;
b) $\frac{3+5 i}{2+6 i}$.
Solution. a) We have

$$
\frac{2+3 i}{5-7 i}=\frac{(2+3 i)(5+7 i)}{(5-7 i)(5+7 i)}
$$

Let's perform the multiplication for the dividend and divisor separately:

$$
\begin{gathered}
(2+3 i)(5+7 i)=10+14 i+15 i+21 i^{2}=-11+29 i ; \\
(5-7 i)(5+7 i)=25-49 i^{2}=25+49=74
\end{gathered}
$$

Thus,

$$
\frac{2+3 i}{5-7 i}=\frac{(2+3 i)(5+7 i)}{(5-7 i)(5+7 i)}=\frac{-11+29 i}{74}=-\frac{11}{74}+\frac{29}{74} i
$$

b) $\frac{3+5 i}{2+6 i}=\frac{(3+5 i)(2-6 i)}{(2+6 i)(2-6 i)}=\frac{6-18 i+10 i-30 i^{2}}{4-36 i^{2}}=\frac{36-8 i}{40}=$ $=\frac{9}{10}-\frac{1}{5} i$.

199-210. Perform the division: