Complex Numbers and Quadratic Equations

m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies

Let z = (3+isinθ)/(4-icosθ) x (4+icosθ)/(4+icosθ)
= (12 - sinθcosθ + i(4sinθ + 3cosθ))/(16 + cos²θ)
z is real
∴ 4sinθ + 3cosθ = 0
⇒ tanθ = -3/4 [? θ lies is 2nd quadrant]
arg(sinθ + icosθ) = π + tan?¹(cosθ/sinθ)
= π - tan?¹(4/3)

p? = α? + β?
= (α + 1)² . α + (β - 1)² . β
= 5α + 5β + 6
= 5(1) + 6 = 11
p? = α² + β² = α + β + 2 = 3
p? = α³ + β³ = (α + 1) . α + (β + 1) . β
= 1 + 3 = 4
Hence p? ≠ p? . p?

α+β = 3/7, αβ = -2/7
α/ (1-α²) + β/ (1-β²) = (α+β) - αβ (α+β) / (1-α²) (1-β²)
= (α+β) - αβ (α+β) / (1 + (αβ)² - (α+β)² - 2αβ)
= (3/7) + (2/7) (3/7) / (1 + (-2/7)² - (3/7)² - 2 (-2/7) = 27/16

(-1+i√3)/ (1-i)³? = (2 (cos (2π/3) + isin (2π/3)/ (√2 (cos (-π/4) - isin (-π/4)³?
= (2³? (cos20π + isin20π)/ (2¹? (cos (-15π/2) - isin (-15π/2)
= (2¹? (1 + 0i)/ (0 + i) = -2¹? i

Re (z²) = x²-y². 2 (Im (z)² = 2y². 2Re (z) = 2x.
x²-y²+2y²+2x=0 ⇒ x²+y²+2x=0.
(x+1)²+y²=1. Center (-1,0).
Parabola: x²-6x+9 = y-13+9 ⇒ (x-3)²=y-4. Vertex (3,4).
Line through (-1,0) and (3,4): slope = (4-0)/ (3- (-1) = 1.
y-0 = 1 (x+1) ⇒ y=x+1.
y-intercept is 1.

x? + 2 (20)¹/? x³ + (20)¹/²x² + . No.
x² = - (20)¹/? x - (5)¹/².
α+β = - (20)¹/? , αβ=5¹/².
α²+β² = (α+β)²-2αβ = (20)¹/² - 2 (5)¹/² = 0.
α? +β? = (α²+β²)² - 2 (αβ)² = 0 - 2 (5) = -10.
α? +β? = (α? +β? )² - 2 (αβ)? = (-10)² - 2 (5)² = 100 - 50 = 50.

S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1

x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48