JEE Mains · Maths · STD 11 - 4.1 complex nubers
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
- A imaginary axis
- B real axis
- C line, \(y=x\)
- D line, \(y=-x\)
Answer & Solution
Correct Answer
(C) line, \(y=x\)
Step-by-step Solution
Detailed explanation
\(z=x+i y\) \(z ^{2}= i | z |^{2}\) \((x+i y)^{2}=i\left(x^{2}+y^{2}\right)\) \(\left(x^{2}-y^{2}\right)-i\left(x^{2}+y^{2}-2 x y\right)=0\) \((x-y)(x+y)-i(x-y)^{2}=0\) \((x-y)((x+y)-i(x-y))=0\) \(\Rightarrow x = y\) z lies on \(y=x\)
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