Let z = x + iy be a non-zero complex number s | vedclass

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(C) Given $z = x + iy$ and $z^{2} = i|z|^{2}.$Substituting $z = x + iy$ and $|z|^{2} = x^{2} + y^{2}$ into the equation:$(x + iy)^{2} = i(x^{2} + y^{2

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line $y = x$

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(C) Given $z = x + iy$ and $z^{2} = i|z|^{2}.$Substituting $z = x + iy$ and $|z|^{2} = x^{2} + y^{2}$ into the equation:$(x + iy)^{2} = i(x^{2} + y^{2})$$x^{2} - y^{2} + 2ixy = i(x^{2} + y^{2})$Equating the real and imaginary parts:Real part: $x^{2} - y^{2} = 0 \Rightarrow (x - y)(x + y) = 0$Imaginary part: $2xy = x^{2} + y^{2}$ $\Rightarrow x^{2} - 2xy + y^{2} = 0$ $\Rightarrow (x - y)^{2} = 0$From the imaginary part,we get $x = y.$Substituting $x = y$ into the real part equation: $y^{2} - y^{2} = 0,$ which is satisfied.Therefore,$z$ lies on the line $y = x.$

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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