Question

z=z₁/z₂+z₃

z₁=-3-2i
z₂=1+9i
z₃=-6+i
z=?

Answer 1

$\displaystyle Z= \frac{z_1}{z_2}+z_3\\\\Z= \frac{-3-2i}{1+9i}+(-6+i)= \frac{(-3-2i)(1-9i)}{(1^2-(9i)^2)}+(-6+i)=\\\\= \frac{-3-2i+27i+18i^2}{1-81i^2}+(-6+i)=\\\\= \frac{-3+25i-18}{1+81}+(-6+i)= \frac{25i-21}{82}+ \frac{82(-6+i)}{82}=\\\\= \frac{25i-21-492+82i}{82}= \frac{107i-513}{82}= \frac{107}{82}i- \frac{513}{82}$