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Question
If |z2 – 1| = |z|2 + 1, then show that z lies on imaginary axis.
Solution
Let z = x + iy,
Then |z2 – 1| = |z|2 + 1
⇒ |x2 – y2 – 1 + i2xy| = |x + iy|2 + 1
⇒ (x2 – y2 – 1)2 + 4x2y2 = (x2 + y2 + 1)2
⇒ 4x2 = 0
i.e., x = 0
Hence z lies on y-axis.
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