Question

Solve the equation $\left|z\right|=z+1+2i$.

Answer 1

Let $z=x+iy$.
Then,
$\left|z\right|=\sqrt{x^2+y^2}$


$$\begin{gathered} \therefore \left|z\right|=z+1+2i \\ \Rightarrow \sqrt{x^2+y^2}=\left(x+iy\right)+1+2i \\ \Rightarrow \sqrt{x^2+y^2}=\left(x+1\right)+i\left(y+2\right) \\ \Rightarrow \sqrt{x^2+y^2}=\left(x+1\right)\mathrm{and}y+2=0 \\ \Rightarrow x^2+y^2={\left(x+1\right)}^2\mathrm{and}y=-2 \\ \Rightarrow x^2+y^2=x^2+1+2x\mathrm{and}y=-2 \\ \Rightarrow y^2=2x+1\mathrm{and}y=-2 \\ \Rightarrow 4=2x+1\mathrm{and}y=-2 \\ \Rightarrow 2x=3\mathrm{and}y=-2 \\ \Rightarrow x=\frac{3}{2}\mathrm{and}y=-2 \end{gathered}$$

$\therefore z=x+iy=\frac{3}{2}-2i$

​Thus, $z=\frac{3}{2}-2i$