Phương trình \(\frac{\left|z\right|^2}{\overline{z}}+iz+\frac{z-i}{1-i}=0\) có nghiệm là
\(z=1-\sqrt{3}i\). \(z=5\). \(z=-21+3\sqrt{2}i\). \(z=\frac{i}{3}\). Hướng dẫn giải:Ta có: \(z.\overline{z}=\left|z\right|^2\) (vì nếu \(z=x+yi\) thì \(\left(x+yi\right)\left(x-yi\right)=x^2+y^2=\left|z\right|^2\))
Vậy: \(\frac{\left|z\right|^2}{\overline{z}}+iz+\frac{z-i}{1-i}=0\)
\(\Leftrightarrow\frac{z.\overline{z}}{\overline{z}}+iz+\frac{\left(z-i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}=0\)
\(\Leftrightarrow z+iz+\frac{z+iz-i-i^2}{1-i^2}=0\)
\(\Leftrightarrow z+iz+\frac{z+iz-i-i^2}{2}=0\)
\(\Leftrightarrow\frac{2z+2iz+z+iz-i+1}{2}=0\)
\(\Leftrightarrow3z+3iz-i+1=0\)
\(\Leftrightarrow3\left(1+i\right)z=i-1\)
\(\Leftrightarrow z=\frac{1}{3}\frac{i-1}{i+1}\)
\(\Leftrightarrow z=\frac{1}{3}\frac{\left(i-1\right)^2}{\left(i+1\right)\left(i-1\right)}\)
\(\Leftrightarrow z=\frac{1}{3}\frac{i^2-2i+1}{i^2-1}\)
\(\Leftrightarrow z=\frac{1}{3}\frac{-2i}{-2}\)
\(\Leftrightarrow z=\frac{i}{3}\).