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If $\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0$, where $z=x+$ iy is a complex number, then which one of the following is correct?
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Verified Answer
The correct answer is:
$|z|=1$
$\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$
$\frac{z-1}{z+1}=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+y^{2}+2 x+1}$
$\Rightarrow \operatorname{Re}\left(\frac{z-1}{z+1}\right)=\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+2 x+1}=0$
$\Rightarrow x^{2}+y^{2}-1=0$
$\Rightarrow x^{2}+y^{2}=1$
Also, $z \bar{z}=x^{2}+y^{2}=1$
and $z \bar{z}=|z|^{2}$
$\Rightarrow|z|^{2}=1$
$\Rightarrow|z|=1$
$\frac{z-1}{z+1}=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+y^{2}+2 x+1}$
$\Rightarrow \operatorname{Re}\left(\frac{z-1}{z+1}\right)=\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+2 x+1}=0$
$\Rightarrow x^{2}+y^{2}-1=0$
$\Rightarrow x^{2}+y^{2}=1$
Also, $z \bar{z}=x^{2}+y^{2}=1$
and $z \bar{z}=|z|^{2}$
$\Rightarrow|z|^{2}=1$
$\Rightarrow|z|=1$
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