Solve this

Solution:

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.