Question:
Let \(z=x+iy\) be a non-zero complex number such that \(z^2=i|z|^2\), where \(i=\sqrt{-1}\), then z lies on the :
Let \(z=x+iy\) be a non-zero complex number such that \(z^2=i|z|^2\), where \(i=\sqrt{-1}\), then z lies on the :
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Expand \(z^2\) and compare real and imaginary parts with \(i(x^2+y^2)\).
Updated On: Nov 5, 2026
- line, \(y=-x\)
- real axis
- line, \(y=x\)
- imaginary axis
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The Correct Option is C
Solution and Explanation
\(z^2=x^2-y^2+2ixy\) and \(i|z|^2=i(x^2+y^2)\).
Real parts: \(x^2-y^2=0 \Rightarrow y=\pm x\).
Imaginary parts: \(2xy=x^2+y^2>0\) (z non-zero), so x and y have the same sign, i.e. \(y=x\).
(Check: \(y=x\) gives \(2x^2=2x^2\), true.) So z lies on the line \(y=x\).
Answer: (3) line, \(y=x\)
Real parts: \(x^2-y^2=0 \Rightarrow y=\pm x\).
Imaginary parts: \(2xy=x^2+y^2>0\) (z non-zero), so x and y have the same sign, i.e. \(y=x\).
(Check: \(y=x\) gives \(2x^2=2x^2\), true.) So z lies on the line \(y=x\).
Answer: (3) line, \(y=x\)
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