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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