Mathematics › Complex Number › Conjugate, modulus and argument
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 Re z denote the real part of z . Let S be the set of all complex numbers z satisfying z 4 - | z | 4 = 4 i z 2 , where i = - 1 . Then the minimum possible value of z 1 - z 2 2 , where z 1 , z 2 ∈ S with Re z 1 > 0 and Re z 2 < 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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