JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 1)
For a complex number z, let Re(z) denote that real part of z. Let S be the set of all complex numbers z satisfying $${z^4} - |z{|^4} = 4i{z^2}$$, where i = $$\sqrt { - 1} $$. Then the minimum possible value of |z1 $$-$$ z2|2, where z1, z2$$ \in $$S with Re(z1) > 0 and Re(z2) < 0 is .........
Answer
8
Explanation
For a complex number z, it is given that,
$${z^4} - |z{|^4} = 4i{z^2}$$
$$ \Rightarrow {z^4} - {z^{2 - 2}}z = 4i{z^2}$$
$$ \Rightarrow {z^2}(z - \overline z )(z + \overline z ) = 4i{z^2}$$
So, either $${z^2} = 0$$ or $$(z - \overline z )(z + \overline z ) = 4i$$
Now, Case - I, if $${z^2} = 0$$ and $$z = x + iy$$
So, $${x^2} - {y^2} + 2ixy = 0$$
$$ \Rightarrow {x^2} - {y^2} = 0$$
and $$xy = 0$$
$$ \Rightarrow x = y = 0$$
$$ \Rightarrow z = 0$$, which is not possible according to given conditions.
Case - II, if $$(z - \overline z )(z + \overline z ) = 4i$$ and
$$z = x + iy$$
So, $$(2iy)(2x) = 4i$$
$$ \Rightarrow $$ xy = 1 is an equation of rectangular hyperbola and for minimum value of $$|{z_1} - {z_2}{|^2}$$, the z1 and z2 must be vertices of the rectangular hyperbola.
Therefore, $${z_1} = 1 + i$$ and $${z_2} = - 1 - i$$
$$ \therefore $$ Minimum value of $$|{z_1} - {z_2}{|^2}$$
$$ = {(1 + 1)^2} + {(1 + 1)^2} = 4 + 4 = 8$$.
$${z^4} - |z{|^4} = 4i{z^2}$$
$$ \Rightarrow {z^4} - {z^{2 - 2}}z = 4i{z^2}$$
$$ \Rightarrow {z^2}(z - \overline z )(z + \overline z ) = 4i{z^2}$$
So, either $${z^2} = 0$$ or $$(z - \overline z )(z + \overline z ) = 4i$$
Now, Case - I, if $${z^2} = 0$$ and $$z = x + iy$$
So, $${x^2} - {y^2} + 2ixy = 0$$
$$ \Rightarrow {x^2} - {y^2} = 0$$
and $$xy = 0$$
$$ \Rightarrow x = y = 0$$
$$ \Rightarrow z = 0$$, which is not possible according to given conditions.
Case - II, if $$(z - \overline z )(z + \overline z ) = 4i$$ and
$$z = x + iy$$
So, $$(2iy)(2x) = 4i$$
$$ \Rightarrow $$ xy = 1 is an equation of rectangular hyperbola and for minimum value of $$|{z_1} - {z_2}{|^2}$$, the z1 and z2 must be vertices of the rectangular hyperbola.
Therefore, $${z_1} = 1 + i$$ and $${z_2} = - 1 - i$$
$$ \therefore $$ Minimum value of $$|{z_1} - {z_2}{|^2}$$
$$ = {(1 + 1)^2} + {(1 + 1)^2} = 4 + 4 = 8$$.
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