Question
If $|\text{z}+1|=\text{z}+2(1+\text{i}),$ find z.

Answer

Let $\text{z}=\text{x}+\text{iy}$
$|\text{z}+1|=\text{z}+2(1+\text{i})$
$\Rightarrow|\text{x}+\text{iy}+1|=\text{x}+\text{iy}+2+2\text{i}$
$\Rightarrow\sqrt{(\text{x}+1)^2+\text{y}^2}=(\text{x}+2)\text{i}(\text{y}+2)$
Comparing, real and imaginary parts, we get
$\text{x}+2=\sqrt{\text{x}^2+2\text{x}+1+\text{y}^2}$ and $\text{y}+2=0$
$\text{y}+2=0$
$\Rightarrow\text{y}=-2$
& $(\text{x}+2)^2=\text{x}^2+2\text{x}+1+\text{y}^2$
$\Rightarrow\text{x}^2+4\text{x}+4=\text{x}^2+2\text{x}+1+\text{y}^2$
$\Rightarrow2\text{x}+3=\text{y}^2$
$\Rightarrow2\text{x}+3=(-2)^2$
$\Rightarrow2\text{x}+3=4$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$
$\therefore\text{z}=\text{x}+\text{iy}=\frac{1}{2}-2\text{i}$

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