Complex NumbersHard
Question
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 zz3 + zz3 = 350 is
Options
A.48
B.32
C.40
D.80
Solution
zz(z2 + z2) = 350
Put z = x + iy
(x2 + y2) (x2 - y2) = 175
(x2 + y2) (x2 - y2) = 5 . 5 . 7
x2 + y2 = 25
x2 - y2 = 7
x = ± 4, y = ± 3
x, y ∈ I
Area = 8 × 6 = 48 sq. unit.
Put z = x + iy
(x2 + y2) (x2 - y2) = 175
(x2 + y2) (x2 - y2) = 5 . 5 . 7
x2 + y2 = 25
x2 - y2 = 7
x = ± 4, y = ± 3
x, y ∈ I
Area = 8 × 6 = 48 sq. unit.
Create a free account to view solution
View Solution FreeMore Complex Numbers Questions
The conjugate of is equal to-...Let S = {z : z = x + iy, y ≥ 0, |z - z0| ≤ 1 }where |z0| = |z0 - ω| = |z0 - ω2|, ωand ω2...If ( − 7 − 24i)1/2 = x − iy, then x2 + y2 is equal to-...The square root of 8 − 6i is -...The principal value of the arg(z) and |z| of the complex number z = 1 + cos + i sin are respectively :...