AP EAMCET20227 Jul 2022Morning ShiftMathematicsComplex NumberActual
Let z=x+i y be a complex number with x, y Z . Then, the area (in sq units) of the rectangle whose vertices are the roots of the equation z z^3+z z ^3=350 is
Options
- A48
- B32
- C40
- D44
Correct answer
A. 48
Step-by-step solution
z z^3+z z ^3=350 z z (z^2+ z ^2 )=350 Let z=x+i y array lc & z =x-i y & (x+i y)(x-i y) [(x+i y)^2+(x-i y)^2 ]=350 array aligned & (x^2+y^2 ) 2 (x^2-y^2 )=350 & (x^2+y^2 ) (x^2-y^2 )=175=25 7 aligned x^2+y^2=25 ...(i) and x^2-y^2=7 ...(ii) On adding Eqs. (i) and (ii), 2 x^2=32 array ll & x^2=16 x= 4 and & y^2=9 y= 3 array Vertices of the rectangle are (4,3),(4,-3),(-4,-3) and (-4,3) . gathered A B=6 and A D=8 Required area =6 8 gathered =48 sq. units