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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


 

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2 Answers

Deepak Kumar
25 Points
7 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
7 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

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