Let z=x+iy be a complex number with x, y in m | vedclass

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(A) Given equation: $\bar{z} z^3+z \bar{z}^3=350$$\Rightarrow z \bar{z}(z^2+\bar{z}^2)=350$Let $z=x+iy$,then $\bar{z}=x-iy$.Substituting these into th

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$48$

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(A) Given equation: $\bar{z} z^3+z \bar{z}^3=350$$\Rightarrow z \bar{z}(z^2+\bar{z}^2)=350$Let $z=x+iy$,then $\bar{z}=x-iy$.Substituting these into the equation:$(x+iy)(x-iy)[(x+iy)^2+(x-iy)^2]=350$$(x^2+y^2)[(x^2-y^2+2ixy)+(x^2-y^2-2ixy)]=350$$(x^2+y^2) \cdot 2(x^2-y^2)=350$$(x^2+y^2)(x^2-y^2)=175$Since $x, y \in \mathbb{Z}$,we have $x^2+y^2=25$ and $x^2-y^2=7$.Adding these equations: $2x^2=32$ $\Rightarrow x^2=16$ $\Rightarrow x=\pm 4$.Subtracting these equations: $2y^2=18$ $\Rightarrow y^2=9$ $\Rightarrow y=\pm 3$.The vertices of the rectangle are $(4,3), (4,-3), (-4,-3),$ and $(-4,3)$.The length of the sides are $6$ and $8$.Area $= 6 	imes 8 = 48$ sq units.

Let $z=x+iy$ be a complex number with $x, y \in \mathbb{Z}$. Then,the area (in sq units) of the rectangle whose vertices are the roots of the equation $\bar{z} \cdot z^3+z \cdot \bar{z}^3=350$ is

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