Question

For a complex number$z$, let $Re\left(z\right)$ denote the 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$|z_1-z_2{|}^2$, where$z_1,z_2\in S$with$Re\left(z_1\right)>0$ and$Re\left(z_2\right)<0$, is _____

Answer 1

Step 1: Simplifying the given equation:

$$\begin{gathered} z^4-\left\{\right|z{|}^2{\}}^2=4i{z}^2 \\ \Rightarrow z^4-\{z\overline{z}{\}}^2=4i{z}^2\left(\because z.\overline{z}=|z{|}^2\right) \\ \Rightarrow z^4-z^2{\left(\overline{z}\right)}^2=4i{z}^2 \\ \Rightarrow z^2[z^2-{\left(\overline{z}\right)}^2]-4i{z}^2=0 \\ \Rightarrow z^2=0or{z}^2-{\left(\overline{z}\right)}^2-4i=0 \\ \Rightarrow z^2=0or{z}^2-{\left(\overline{z}\right)}^2=4i \end{gathered}$$

Step 2: Put$z=x+iy$ in the above expression:

$$\begin{gathered} z^2-{\left(\overline{z}\right)}^2=4i \\ \Rightarrow {(x+iy)}^2-{(x-iy)}^2=4i \\ \Rightarrow 4xyi=4i \\ \Rightarrow xy=1 \end{gathered}$$

Step 3: Find the simplified expression of$|z_1-z_2{|}^2$:

$$\begin{gathered} |z_1-z_2{|}^2=(z_1-z_2)(\overline{z_1-z_2}) \\ \Rightarrow |z_1-z_2{|}^2={(x_1-x_2)}^2+{(y_1-y_2)}^2 \\ \Rightarrow |z_1-z_2{|}^2=x_1^2+x_2^2-2x_1{x}_2+y_1^2+y_2^2-2y_1{y}_2 \end{gathered}$$

Step 4: Apply A.M and G.M to the above numbers:

Consider$x_1^2,x_2^2,y_1^2,y_2^2,x_1(-x_2),x_1(-x_2),y_1(-y_2)\&y_1(-y_2)$ are different numbers, then A.M and G.M of these numbers are as follows,

$$\begin{gathered} \Rightarrow \frac{x_1^2+x_2^2+y_1^2+y_2^2+x_1(-x_2)+x_1(-x_2)+y_1(-y_2)+y_1(-y_2)}{8}\ge \sqrt[8]{x_1^2.x_2^2.y_1^2.y_2^2.x_1(-x_2).x_1(-x_2).y_1(-y_2).y_1(-y_2)} \\ \Rightarrow \frac{x_1^2+x_2^2-2x_1{x}_2+y_1^2+y_2^2-2y_1{y}_2}{8}\ge \sqrt[8]{x_1^2.x_2^2.y_1^2.y_2^2.x_1^2.x_2^2.y_1^2.y_2^2} \\ \Rightarrow |z_1-z_2{|}^2\ge 8{\left(1\right)}^{\frac{1}{8}}\because x_1{y}_1=1\&x_2{y}_2=1 \end{gathered}$$

Hence, the minimum value of$\mathbf{|}{\mathit{z}}_{\mathbf{1}}\mathbf{-}{\mathit{z}}_{\mathbf{2}}{\mathbf{|}}^{\mathbf{2}}$ is $\mathbf{8}$