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

Let z = x + iy

Consider, |z + i| = |z – i|

⇒ |x + iy + i| = |x + iy – i|

⇒ |x + i(y +1)| = |x + i(y – 1)|

Squaring both the sides, we get

⇒ x² + y² + 1 + 2y = x² + y² + 1 – 2y

⇒ x² + y² + 1 + 2y – x² – y² – 1 + 2y = 0

⇒ 2y + 2y = 0

⇒ 4y = 0

⇒ y = 0

Putting the value of y in eq. (i), we get

z = x + i(0)

⇒ z = x

Hence, z is purely real.