For a complex number z , let Re z denote the real part of z . Let S be the set of all complex numbers z…

For a complex number z, let Rez denote the real part of z. Let S be the set of all complex numbers z satisfying z4-|z|4=4iz2, where i=-1. Then the minimum possible value of z1-z22, where z1,z2S with Rez1>0 and Rez2<0, is _______

Solution

$z^4 - |z|^4 = 4iz^2 \Rightarrow z^4 - |z|^2 = 4iz^2$ $\Rightarrow z^2(z^2 - |z|^2) = 4iz^2 \Rightarrow z^2 - |z|^2 = 4i$ $\Rightarrow (z + \bar{z})(z - \bar{z}) = 4i$ $\Rightarrow \frac{z + \bar{z}}{2} \frac{z - \bar{z}}{2i} = 1 \Rightarrow xy = 1$ for $z_1$ and $z_2 \Rightarrow x_1y_1 = 1$ and $x_2y_2 = 1$ $x_1$ and $x_2$ are of opposite sign, similarly $y_1$ and $y_2$ are of opposite sign $\Rightarrow x_1 > 0, y_1 > 0, x_2 < 0, y_2 < 0$ Now $|z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$ $= x_1^2 + x_2^2 + y_1^2 + y_2^2 - 2x_1x_2 - 2y_1y_2$ $= x_1^2 + x_2^2 + y_1^2 + y_2^2 + x_1(-x_2) + x_1(-x_2) + y_1(-y_2) + y_1(-y_2)$ $\geq 8(x_1^2 \cdot x_2^2 \cdot y_1^2 \cdot y_2^2 \cdot x_1(-x_2) \cdot x_1(-x_2) \cdot y_1(-y_2) \cdot y_1(-y_2))^{1/8}$ $\geq 8((x_1y_1)^4 \cdot (x_2y_2)^4)^{1/8}$ $\geq 8$.

Asked in: JEE Advanced 2020 (Paper 2)

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