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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $z\bar{z}^3+\bar{z}z^3=350$.We can factor this as $z\bar{z}(z^2+\bar{z}^2)=350$.Let $z=x+iy$,then $z\bar{z}=x^2+y^2$ and $z^2+\

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Given the equation $z\bar{z}^3+\bar{z}z^3=350$.We can factor this as $z\bar{z}(z^2+\bar{z}^2)=350$.Let $z=x+iy$,then $z\bar{z}=x^2+y^2$ and $z^2+\bar{z}^2=(x+iy)^2+(x-iy)^2=2(x^2-y^2)$.Substituting these into the equation: $(x^2+y^2)(2(x^2-y^2))=350$.$(x^2+y^2)(x^2-y^2)=175$.Since $175 = 25 	imes 7$,we set $x^2+y^2=25$ and $x^2-y^2=7$.Adding these equations gives $2x^2=32$,so $x^2=16$,which means $x=\pm 4$.Subtracting gives $2y^2=18$,so $y^2=9$,which means $y=\pm 3$.The vertices of the rectangle are $(4,3), (4,-3), (-4,-3),$ and $(-4,3)$.The length of the rectangle is $|4 - (-4)| = 8$ and the breadth is $|3 - (-3)| = 6$.Area $= 8 	imes 6 = 48$ square units.

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

Similar Questions

Which of the following statements is correct?

When $n-p-n$ transistor is used as an amplifier:

The common Indian bullfrog is:

$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is

The median of a set of $9$ distinct observations is $20.5$. If each of the largest $4$ observations of the set is increased by $2$,then the median of the new set

Vedclass Test Series

Exam Paper Generator

Online Exam Module