Let z=x+iy be a complex number,where x and y | vedclass

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(A) Given,$z\bar{z}^3+\bar{z}z^3=350$ $\Rightarrow z\bar{z}(\bar{z}^2+z^2)=350$ $\Rightarrow |z|^2(x-iy)^2+(x+iy)^2=350$ $\Rightarrow (x^2+y^2)(x^2-y^

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$48$

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(A) Given,$z\bar{z}^3+\bar{z}z^3=350$ $\Rightarrow z\bar{z}(\bar{z}^2+z^2)=350$ $\Rightarrow |z|^2(x-iy)^2+(x+iy)^2=350$ $\Rightarrow (x^2+y^2)(x^2-y^2-2ixy+x^2-y^2+2ixy)=350$ $\Rightarrow (x^2+y^2)(2x^2-2y^2)=350$ $\Rightarrow 2(x^2+y^2)(x^2-y^2)=350$ $\Rightarrow x^4-y^4=175$ Since $x$ and $y$ are integers,we test values: $x^4-y^4=(x^2-y^2)(x^2+y^2)=175$. For $x=4, y=3$: $4^4-3^4=256-81=175$. Thus,the vertices are $(4,3), (-4,3), (-4,-3), (4,-3)$. The length of the rectangle is $|4-(-4)|=8$ and the breadth is $|3-(-3)|=6$. Area $= 8 	imes 6 = 48 	ext{ sq. units}$.

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

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