Solve this


If $|z+i|=|z-i|$, prove that $z$ is real.



Let z = x + iy

Consider, $|z+i|=|z-i|$

$\Rightarrow|x+i y+i|=|x+i y-i|$


$\Rightarrow \sqrt{(x)^{2}+(y+1)^{2}}=\sqrt{(x)^{2}+(y-1)^{2}}$

$\left[\because|\mathrm{z}|=\right.$ 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


$\Rightarrow Z=X$

Hence, z is purely real.


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