Question

For a complex number $z$, let $\operatorname{Re} \left( z \right)$denote the real part of $z$. Let $S$ be the set of all complex numbers $z$ satisfying ${z^4} - {\left| z \right|^4} = 4i{z^2}$, where $i = \sqrt { - 1} $. Then the minimum possible value of ${\left| {{z_1} - {z_2}} \right|^2}$, where ${z_1},{z_2} \in S$ with $\operatorname{Re} \left( {{z_1}} \right) > 0$ and $\operatorname{Re} \left( {{z_2}} \right) < 0$, is _________.

Answer 1

Hint: In this question, we are given that $S$ is the set of all complex numbers of $z$ which satisfies ${z^4} - {\left| z \right|^4} = 4i{z^2}$. Let, $z = x + iy$ and use the formula ${\left| z \right|^2} = z\overline z $ and solve further. Now to calculate the minimum value of ${\left| {{z_1} - {z_2}} \right|^2}$, apply the formula \[{\left| {{z_1} - {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2\operatorname{Re} \left( {{z_1}\overline {{z_2}} } \right)\]. In last use the condition that arithmetic mean is greater than or equal to geometric mean and solve.

Formula Used:
Complex number formulas –
$z = x + iy$
Conjugate $\overline z = x - iy$
${\left| z \right|^2} = z\overline z $
\[{\left| {{z_1} - {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2\operatorname{Re} \left( {{z_1}\overline {{z_2}} } \right)\]

Complete step by step Solution:
Let, the complex number $z$be $x + iy$,
It means \[{z_1} = {x_1} + i{y_1},{z_2} = {x_2} + i{y_2}\]
The conjugate of $z$ will be $\overline z = x - iy$

Given that,
${z^4} - {\left| z \right|^4} = 4i{z^2}$
${z^4} - {\left( {z\overline z } \right)^2} = 4i{z^2}$
${z^2}\left( {{z^2} - {{\left( {\overline z } \right)}^2}} \right) = 4i{z^2}$
${z^2} - {\left( {\overline z } \right)^2} = 4i$
${\left( {x + iy} \right)^2} - {\left( {x - iy} \right)^2} = 4i$
${x^2} + {i^2}{y^2} + 2ixy - \left( {{x^2} + {i^2}{y^2} - 2ixy} \right) = 4i$
${x^2} + {i^2}{y^2} + 2ixy - {x^2} - {i^2}{y^2} + 2ixy = 4i$
\[4ixy = 4i\]
\[xy = 1\]
Which implies that, \[{x_1}{y_1} = 1,{x_2}{y_2} = 1 - - - - - - \left( 1 \right)\]

Also given, \[\left\{ \begin{gathered}
  {z_1} = {x_1} + i{y_1} \\
  {z_2} = {x_2} + i{y_2} \\
  {x_1} > 0,{x_2} < 0 \\
  {y_1} > 0,{y_2} < 0 \\
\end{gathered} \right.\]

Now, as we know that \[{\left| {{z_1} - {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} - 2\operatorname{Re} \left( {{z_1}\overline {{z_2}} } \right)\]
\[ \Rightarrow {\left| {{z_1} - {z_2}} \right|^2} = {x_1}^2 + {x_2}^2 - {x_1}{x_2} - {x_1}{x_2} + {y_1}^2 + {y_2}^2 - {y_1}{y_2} - {y_1}{y_2}\]

Using the condition, \[A.M \geqslant G.M\] (A.M., G.M. are arithmetic mean and geometric mean), Each successive term in an arithmetic progression is obtained by adding the common difference to the preceding term. Each successive term in a geometric progression is obtained by multiplying the common ratio by the preceding term.

\[\frac{{{x_1}^2 + {x_2}^2 - {x_1}{x_2} - {x_1}{x_2} + {y_1}^2 + {y_2}^2 - {y_1}{y_2} - {y_1}{y_2}}}{8} \geqslant {\left( {{x_1}^2{x_2}^2{y_1}^2{y_2}^2{x_1}^2{x_2}^2{y_1}^2{y_2}^2} \right)^{\frac{1}{8}}}\]

From equation (1),
\[\frac{{{x_1}^2 + {x_2}^2 - {x_1}{x_2} - {x_1}{x_2} + {y_1}^2 + {y_2}^2 - {y_1}{y_2} - {y_1}{y_2}}}{8} \geqslant {\left( 1 \right)^{\frac{1}{8}}}\]
\[{\left| {{z_1} - {z_2}} \right|^2} \geqslant 8\]

Hence, the minimum possible value of ${\left| {{z_1} - {z_2}} \right|^2}$, where ${z_1},{z_2} \in S$ with $\operatorname{Re} \left( {{z_1}} \right) > 0$and $\operatorname{Re} \left( {{z_2}} \right) < 0$, is $8$.

Note: The key concept involved in solving this problem is a good knowledge of arithmetic and geometric mean. Students must remember that arithmetic mean is the average of all the numbers in the series whose sum is divided by the total count of the numbers in the series. On the other hand, the Geometric mean is the compounding effect of the numbers in a series multiplied by the $nth$ root of the multiplication.