Test:- Complex Number - 2 - Mathematics MCQ

 , it is bisector of line segment joining (-3.,0) and (5,0)

I z - 4 I = | z + 4|
If ( z - 4 ) = k |z + 4| k ≠ 1
(x - y)² + y² = ± (x+y)² + y² then it is circle
but now x = 0
x = 0 hence it is y - axis.

it is a circle

 | z - 4 | < | z - 2 |
=> | x - 4 + iy | < | x + 2 - iy |
=> (x - y)² + y² < (x - 2)² + y²
=> x²- 8 x + 16 < x² - 4 x +4
=> - 4 x < -12
=> x > 3
=> real value of z should be > 3

The concept of inequality does not hold in the set of complex number 

 Let. z = x + iy

and |w| = 1
 
squaring both side

Hence z = x + iy = x + 0
which is real

z₁, z₂, z₃, are in AP, so they lie on a straight line (by mid point method)

 is
= -i (sum of complex roots of 1)
= i

z₁, z₂, z₃, z₄, are the complex numbers which are the vertices of a parallelogram taken in order if

Diagonal of a parelielogam bisect each other at midpoint of AC and BD
∴ 
z₁ + z₃ = z₂ + z₄

 1, ω, ω² are cube roots of one (unity).

If a and b are cube roots of unity
x = a + b , y = ωa + bω²

If 1, ω, ω²..., ω^n-1 are the n^th root of unity then

if ω₁, ω₂, are the complex roots of unity

= (ω²) + (ω)

1 ,ω, ω² are root of cube root of unity , then
(x - 1)³ + 8 = 0


hence required roots are - 1 , 1 - 2w , 1 - 2w

Value of |z| + |z - 1| is at least |1 - 0 |
so minimum value of |z| + |z - 1| is 1

tan θ = 1/1
θ = π/4

Area, AABC =

As e^π/2 is i so, iz is obtained by rotating z by an angle of π/2. so , angle between z and iz is π/2.

A, B, C are equilateral A and P be the centroid of ΔABC.

since AB = BC = CA = 2a
Δ AOC
sin 60 = AO/CA