Phương trình \(\left(\left(3+i\right)\overline{z}+2+2i\right)\left(iz+\frac{2+i}{i}\right)=0\) có nghiệm là
\(z=\frac{-4-2i}{5},z=2+i.\) \(z=\frac{-4-5i}{5};z=2i\). \(z=\frac{-4+2i}{5};z=2+i.\) \(z=\frac{-4+2i}{5};z=2i\). Hướng dẫn giải:\(\left(\left(3+i\right)\overline{z}+2+2i\right)\left(iz+\frac{2+i}{i}\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}\left(3+i\right)\overline{z}+2+2i=0\\iz+\frac{2+i}{i}=0\end{array}\right.\)
Trường hợp 1: \(\left(3+i\right)\overline{z}+2+2i=0\Leftrightarrow\overline{z}=-\frac{2+2i}{3+i}=-\frac{\left(2+2i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}=-\frac{8+4i}{10}=\frac{-4-2i}{5}\)
suy ra \(z=\frac{-4+2i}{5}\).
Trường hợp 2: \(iz+\frac{2+i}{i}=0\Leftrightarrow z=-\frac{2+i}{i^2}=2+i\).