0

Guest

Courses New

z=x+iy,x,y:z, then the area of the rectangle whose vertices are the roots of z(conjugate of z)^3+(conjugate of z)z^3=1342, is

z=x+iy,x,y:z, then the area of the rectangle whose vertices are the roots of  z(conjugate of z)^3+(conjugate of z)z^3=1342, is


 

Grade:

2 Answers

Deepak Kumar
25 Points
10 years ago

Let Z be the complex number and z be its conjugate.

Now, Z*z^3 + Z^3*z = 1342

or, |Z|^2*{Z^2+z^2} = 1342 (Z*z = |Z|^2)

or, (x^2+y^2)*2(x^2-y^2) = 1342

or, (x^4-y^4) = 671

 

Clearly x = 6 and y = 5 satisfies the equation.

Hence, Z = 6+5i, 6-5i, -6+5i and -6-5i are the vertices of the rectangle.

And the area is (2*6)*(2*5) = 120 sq. units.

 

Thanks and Regards,
Deepak Kumar
AskIITians Faculty,
B. Tech, IIT Delhi.

shubham sharma
27 Points
10 years ago

Zz^3+zZ^3=1342. {Z=x+iy. z=conjugate of z just for simplification}

Zz(z^2)+zZ(Z^2)=1342

Mod(Z)^2[z^2+Z^2)=1342. {since zZ=modZ}

Now, x^2+y^2(2x^2-2y^2)=1342. {mod(Z) = underroot(x^2+y^2)

Now [x^2+y^2][x^2-y^2]=671

Solve for x and y by taking two multiples of 671

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More