Let z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex number.Further,assume that the origin , z1 and z2 form an equilateral triangle.Then
a2 = b
a2 = 2b
a2 = 3b
a2 = 4b
If and | ω | = 1, then z lies on
A circle
An ellipse
A parabola
A straight line
(x = 3, y = 1 )
(x = 1, y = 3 )
(x = 0 , y = 0 )
The solution of the equation | z | - z = 1 + 2i is
2 - 3/2 i
3/2 + 2i
3/2 - 2i
-2 + 3/2 i
The complex numbers z1,z2 and z3 satisfying are the vertices of a triangle, which is
Of zero area
Equilateral
Right - angled isosceles
Obtuse - angled isosceles
If the amplitude of a complex number is π/2, then the number is
Purely imaginary
Purely real
Neither real nor imaginary
If -i + 3 is a root of x2 - 6x + k = 0 then the value of k is
5
√5
√10
10
If 1, a1 , a2 , a3 . . . an-1 are the nth roots of unity , then the value of (1 - a1) (1 - a2) (1 - a3) . . . (1 - an-1) is
0
1
n
-n
If z = 1 + i , then the multiplicative inverse of z2 is (where i = √-1 )
2 i
p>1 - i
-i/2
i/2
If z1,z2 and z3 be the vertices of an equilateral triangle occurring in anticlockwise sense, then
z1 + z2 + z32 = 0
z1z2 + z2z3 + z3z1 = 0
z1 + ω z2 + ω2 z3 = 0
z12 + z22 + z32 = 2 (z1z2 + z2z3 + z3z1)
