Question:
Solve the equation $|z|=z+1+2 i$
Solution:
Let $z=x+i y$.
Then,
$|z|=\sqrt{x^{2}+y^{2}}$
$\therefore|z|=z+1+2 i$
$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+i y)+1+2 i$
$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)+i(y+2)$
$\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)$ and $y+2=0$
$\Rightarrow x^{2}+y^{2}=(x+1)^{2}$ and $y=-2$
$\Rightarrow x^{2}+y^{2}=x^{2}+1+2 x$ and $y=-2$
$\Rightarrow y^{2}=2 x+1$ and $y=-2$
$\Rightarrow 4=2 x+1$ and $y=-2$
$\Rightarrow 2 x=3$ and $y=-2$
$\Rightarrow x=\frac{3}{2}$ and $y=-2$
$\therefore z=x+i y=\frac{3}{2}-2 i$
Thus, $z=\frac{3}{2}-2 i$
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