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.

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