If $|z+i|=|z-i|$, prove that $z$ is real.
Let z = x + iy
Consider, $|z+i|=|z-i|$
$\Rightarrow|x+i y+i|=|x+i y-i|$
$\Rightarrow|x+i(y+1)|=|x+i(y-1)|$
$\Rightarrow \sqrt{(x)^{2}+(y+1)^{2}}=\sqrt{(x)^{2}+(y-1)^{2}}$
$\left[\because|\mathrm{z}|=\right.$ modulus $\left.=\sqrt{a^{2}+b^{2}}\right]$
$\Rightarrow \sqrt{x^{2}+y^{2}+1+2 y}=\sqrt{x^{2}+y^{2}+1-2 y}$
Squaring both the sides, we get
$\Rightarrow x^{2}+y^{2}+1+2 y=x^{2}+y^{2}+1-2 y$
$\Rightarrow x^{2}+y^{2}+1+2 y-x^{2}-y^{2}-1+2 y=0$
$\Rightarrow 2 y+2 y=0$
$\Rightarrow 4 y=0$
$\Rightarrow y=0$
Putting the value of y in eq. (i), we get
$z=x+i(0)$
$\Rightarrow Z=X$
Hence, z is purely real.
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