Question

Let $z=x+iy$ be a complex where x and y are integer. Then the area of the rectangle whose vertices are the roots of the equation $\overline{z}{z}^3+z{\overline{z}}^3=$ 350 is

• A. 48
• B. 32
• C. 40
• D. 80

Answer 1

Answer: (A)

48

Given,

$z=x+iy\&x$ and $y$ are integers.

To find are a of rectangle whose vertices are

the roots of the equation.

$\overline{z}{z}^3+z{\overline{z}}^3=350$

Solution,

From given equation.

$\overline{z}{z}^3+z{\overline{z}}^3=350$

$z\overline{z}(\overline{z}{)}^2+\overline{z}z(\overline{z}{)}^2=350z\overline{z}(z^2+{\overline{z}}^2)=350$

We know that $z=x+iy$ than $\overline{z}=x-iy$

$\therefore (x+iy)(x-iy)\left[{(x+iy)}^2{(x-iy)}^2\right]=350[x^2+i^2{(-y)}^2][x^2+i^2{y}^2+2xyi{+x}^2+i^2{y}^2-2xyi]=350(x^2+y^2)({2x}^2-{2y}^2)=350\{i^2=-1\}(x^2+y^2)(x^2-y^2)=175$

If we consider then the following solution can be

predicted $x=\pm 4$ $y=\pm 3$

So, points/vertices of rectangle are

$(4,3),(4,-3),(-4,-3),(-4,3)$

Hence,length of rectangle be$\sqrt{{(-3+3)}^2+{(-4-4)}^2}$

length$=8unit$

${\parallel }^{rly}breadth=6unit$

Hence, area of rectangle be= length $\times$breadth

$=8\times 6$

$=48units.$