Question
Solve the equation $|\text{z}|=\text{z}+1+2\text{i}.$

Answer

Let $\text{z}=\text{x}+\text{iy}.$
Then,
$|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
$\therefore|\text{z}|=\text{z}+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+\text{iy})+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)+\text{i}(\text{y}+2)$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)$ and $\text{y}+2=0$
${\text{x}^2+\text{y}^2}=(\text{x}+1)^2$ and $\text{y}=-2$
${\text{x}^2+\text{y}^2}=\text{x}^2+1+2\text{x}$ and $\text{y}=-2$
$\Rightarrow\text{y}^2=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow4=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=3$ and $\text{y}=-2$
$\Rightarrow\text{x}=\frac{3}{2}$ and $\text{y}=-2$
$\therefore\text{z}=\text{x}+\text{iy}=\frac{3}{2}-2\text{i}$

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