Complex Numbers & Quadratic Equations- 2 - JEE MCQ

α = |z/z¯|
z = x + iy
z¯ = x - iy
|α| = |z/z¯|
⇒ [(x² + y²)^1/2]/[[x² + y²)^1/2]
⇒ 1

z₁ = (√3/2 + i/2), z₂ = (√3 - i/2)
(z1)³ = (√3/2 + i/2)² 
= 3/4 - 1/4 + √3i/2 
= 1/2 + √3i/2
(z₃)³ = (z₁)² * z₁ 
= (1/2 + √3i/2) * (√3/2 + i/2)
= √3/4 + i/4 + 3i/4 - √3/4
= i
(z₂)² = (√3/2 - i/2)² 
= 3/4 - 1/4 - √3i/2 
= 1/2 - √3i/2
(z₂)³ = (z₂)² * z₂
= (1/2 - √3i/2) * (√3/2 - i/2)
= √3/4 - i/4 - 3i/4 - √3/4 
= -i
(z₂)⁵ = (z₂)³ * (z₂)² 
= (-i) * (1/2 - √3i/2)
= -i/2 + √3/2
z = (z₁)⁵ + (z₂)⁵
= i/2 - √3/2 - i/2 + √3/2
= 0
So Im(z) = 0

z = x + iy
z = -x + 0i {-x > 0}
Principle arg(x) = π

Let z = x + iy
Then z² =  x² + 2ixy - y²
And real(z2) = x² - y²  = 0  …………...(1)
And |z| = √x² + y²
|z| = 2
Hence √x² + y² = 2  ………..(2) 
Or x² +  y² = 4 …….(3)
Solving (1) and (3) , we get x = ± √2 and 
y = ± √2
It has 4 solutions.

 z = x+iy
Hence ,(x−2) + i(y−3) = λ
Since the argument is π/4
​⇒ (x−2) = (y−3)
⇒ x − y = −1
⇒ y = x + 1
⇒ x - y + 1 = 0
Thus, slope is positive.

z₁ = |z₁(cosθ₁ + i sinθ₁)
z₂ = |z₂(cosθ₂ + i sinθ₂)|
Given that |z₁| = |z₂|
And arg(z₁) + arg(z₂) = pi
θ₁ + θ₂ = pi
⇒ θ₁ = pi - θ²
Now, z₁ = |z2|[cosθ(pi - θ₂) + i sin(pi - θ₂)]
= z₁ = z₂(-cosθ₂ + i sinθ₂)
= z₁ = - |z₂|(cosθ₂ - i sinθ₂)|
= z₁ = - |z₂|(cosθ₂ - i sinθ₂)
⇒ z₁ = - z₂(bar)

The given expression can be simplified as follows:

-1 + √(-3) = -1 + i√3
-1 - √(-3) = -1 - i√3

Now, (-1 + i√3)² = (-1)² + 2*(-1)*(i√3) + (i√3)²
= 1 - 2i√3 - 3
= -2i√3 - 2

Similarly, (-1 - i√3)² = (-1)² + 2*(-1)*(-i√3) + (-i√3)²
= 1 + 2i√3 - 3
= 2i√3 - 2

Adding the two results together:
(-2i√3 - 2) + (2i√3 - 2) = -4

Therefore, the value of the expression is -4.

Since α³ +1=(α+1)(α² −α+1), using the given information, we get α³ +1 = 0

z = x + iy
|z| = (x² + y²)^1/2
x² + y² = 1
Equation of circle with centre (0,0) and radius 1 unit.

As b and c are linear constants ,independent of x and y, then by substituting them in the equation given, we get an equation linear in x and y. Thus, the given equation represents a straight line.

sin(6π/5) + i(1+cos(6π/5))  or −sin(π/5) + i(1−cos(π/5)) lies in the second quadrant of complex plane hence its argument is given as
arg(x+iy) = π − tan^-1 |y/x| (∀ x<0,y≥0)
= π−tan^-1 |1−cos(π/5)/sin(π/5)|
= π−tan^-1  |2sin2(π/10)/2sin(π/10)cos(π/10)|
= π−tan^-1 |sin(π/10)/cos(π/10)|
= π − tan^-1 |tan(π/10)|
= π−tan^-1 (tan(π/10))  (∵ tan(π/10)>0)
= π−π/10(∵−π/2≤tan−1(x)≤π/2)
= 9π/10
 

(z+1) (z(bar)+1)
((x+1)+iy) ((x+1)-iy)
= (x+1)² + y²
= |(x+1)² + iy²|
= |(x+iy) + 1|²
= |z+1|²