Complex Numbers and Quadratic Equations

The value of ((1+sin(2π/9)+icos(2π/9))/(1+sin(2π/9)-icos(2π/9)))³ is

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The value of (1+sin (2π/9)+icos (2π/9)/ (1+sin (2π/9)-icos (2π/9)³
= (1+cos (5π/18)+isin (5π/18)/ (1+cos (5π/18)-isin (5π/18)³
= (2cos² (5π/36)+2isin (5π/36)cos (5π/36)/ (2cos² (5π/36)-2isin (5π/36)cos (5π/36)³
= (cos (5π/36)+isin (5π/36)/ (cos (5π/36)-isin (5π/36)³
= (e^ (i5π/36)/e^ (-i5π/36)³ = (e^ (i5π/18)³ = e^ (i5π/6) = cos (5π/6)+isin (5π/6)
= -√3/2 + i/2

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a month ago

A ray of light coming from the point (2,2√3) is incident at an angle 30° on the line x=1 at the point A. The ray gets reflected on the line x=1 and meets x-axis at the point B. Then, the line AB passes through the point:

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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

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a month ago

If $\frac{3 + i s i n θ}{4 - i c o s θ}, θ \in [0,2 π]$ , is a real number, then an argument of $s i n ? θ + i c o s ? θ$ is:

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Contributor-Level 10

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)

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a month ago

Let $α$ and $β$ be the roots of the equation $x^{2} - x - 1 = 0$ . If $P_{k} = (α)^{k} + (β)^{k}, k \geq 1$ , then which one of the following statements is not true?

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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?

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New answer posted

a month ago

If α and β are the roots of the equation, 7x²-3x-2=0, then the value of α/(1-α²) + β/(1-β²) is equal to:

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Contributor-Level 10

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

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a month ago

The value of ((-1+i√3)/(1-i))³⁰ is:

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Contributor-Level 10

(-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

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a month ago

The equation of a circle is Re(z²) + 2(Im(z))² + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x² - 6x - y + 13 = 0, has y-intercept equal to……….

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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.

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New answer posted

a month ago

Let α, β be two roots of the equation x² + (20)¹/⁴x + (5)¹/² = 0 Then α⁸ + β⁸ is equal to:

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Contributor-Level 10

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.

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a month ago

Let C be the set of all complex numbers. Let
S₁ = {z ∈ C | |z - 3 - 2i|² = 8},
S₂ = {z ∈ C | Re(z) ≥ 5} and
S₃ = {z ∈ C | |z - z̄| ≥ 8}.
Then the number of element in S₁ ∩ S₂ ∩ S₃ is equal to:

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Contributor-Level 10

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

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New answer posted

a month ago

The number of real solutions of the equation, x² - |x| - 12 = 0 is:

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