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.