If we express ( 2 + 3 i ) 2 in the form of ( x + iy ) , we get
-5 + 12 i
12 - 5 i
5 - 12 i
12 + 5 i
If -i + 3 is a root of x2 - 6x + k = 0 then the value of k is
5
√5
√10
10
(x = 3, y = 1 )
(x = 1, y = 3 )
(x = 0 , y = 0 )
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
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 and | ω | = 1, then z lies on
A circle
An ellipse
A parabola
A straight line
a = 0 and b = 1
a = 1 and b = 0
a = 2 and b = -1
a = -1 and b = 2
If z = 1 + i , then the multiplicative inverse of z2 is (where i = √-1 )
2 i
p>1 - i
-i/2
i/2
If P is a multiple of n , then the sum of Pth power of nth roots of unity is
P
None of these
