Question:
Modulus of z square -1=modulus of z square 1
Answer:
Let z = x + iy
Given |z2 - 1| = |z|2 + 1
=> |(x + iy)2 - 1| = |(x + iy)|2 + 1
=> |(x2 - y2 - 1) + 2ixy| = x2 - y2 + 1
=> (x2 - y2 - 1)2 + 4x2 *y2 = (x2 - y2 + 1)2
=> x = 0
So, z lies on the imaginary axis
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