Let $\mathrm{z}=x+\mathrm{i} y$ be a complex number, where $x$ and $y$ are integers and $\mathrm{i}=\sqrt{-1}$. Then the area of the rectangle whose vertices are the roots of the equation $z^3+\bar{z} z^3=350$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } \mathrm{z}(\overline{\mathrm{z}})^3+\overline{\mathrm{z}} \mathrm{z}^3=350 \ & \Rightarrow \mathrm{z} \overline{\mathrm{z}}(\overline{\mathrm{z}})^2+\overline{\mathrm{z}} \mathrm{zz}^2=350\end{aligned}$ $\begin{aligned} & \Rightarrow|z|^2\left\{(\bar{z})^2+z^2ight\}=350 \quad \ldots\left[\because \bar{z}=|z|^2ight] \ & \Rightarrow|z|^2\left[(x-\mathrm{i} y)^2+(x+\mathrm{i} y)^2ight]=350 \ & \Rightarrow 2\left(x^2+y^2ight)\left(x^2-y^2ight)=350 \ & \Rightarrow 2\left(x^4-y^4ight)=350 \ & \Rightarrow x^4-y^4=175 \ & \Rightarrow x^4=256, y^4=81 \quad \ldots[\because x, y 	ext { are integers }] \ & \Rightarrow x^2=16, y^2=9 \ & \Rightarrow x= \pm 4, y= \pm 3 \end{aligned}$ $	herefore \quad$ vertices of the rectangle are $(4,3) ;(-4,3)$, $(-4,-3)$ and $(4,-3)$. $\begin{aligned} 	herefore \quad 	ext { Required area } & =	ext { length } 	imes 	ext { breadth } \ & =8 	imes 6 \ & =48 	ext { sq. units }\end{aligned}$

Let $\mathrm{z}=x+\mathrm{i} y$ be a complex number, where $x$ and $y$ are integers and…

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