Complex Numbers and Quadratic Equations
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Let α and β be two real roots of the equation (k + 1)tan²x − √2 ⋅ λtanx = (1 – k), where k(≠ -1) and λ are real numbers. If tan²(α + β) = 50, then a value of λ is:
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Let α and β be two real roots of the equation (k + 1)tan²x − √2 ⋅ λtanx = (1 – k), where k(≠ -1) and λ are real numbers. If tan²(α + β) = 50, then a value of λ is:
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For the quadratic equation (k+1)tan²x - √2λ tanx + (k-1) = 0, the roots are tanα and tanβ.
Sum of roots: tanα + tanβ = √2λ / (k+1).
Product of roots: tanα tanβ = (k-1) / (k+1).
tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)
tan (α + β) = [√2λ / (k+1)] / [1 - (k-1)/ (k+1)]
tan (α + β) = [√2λ / (k+1)] / [ (
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If Re((z-1)/(2z+i)) = 1, where z = x + iy, then the point (x, y) lies on a:
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If Re((z-1)/(2z+i)) = 1, where z = x + iy, then the point (x, y) lies on a:
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Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part o
Upvote Let a complex number z, |z| ≥ 1, satisfy log?/√² (|z|+11)/(|z|-1)² ≤ 2. Then, the largest value of |z| is equal to…
Let a complex number z, |z| ≥ 1, satisfy log?/√² (|z|+11)/(|z|-1)² ≤ 2. Then, the largest value of |z| is equal to…
log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0
⇒ |z| - 7 ≤ 0
∴ |z|max = 7
Upvote The least value of |z| where z is complex number which satisfies the inequality exp( ((|z|+3)(|z|-1))/(|z|+1) logₑ2 ) ≥ log_√2 |5√7 + 9i|, i = √-1, is equal to :
The least value of |z| where z is complex number which satisfies the inequality exp( ((|z|+3)(|z|-1))/(|z|+1) logₑ2 ) ≥ log_√2 |5√7 + 9i|, i = √-1, is equal to :
The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3)
Upvote If the integral ∫[0 to 10] (sin(2πx)) / (e^(x-[x])) dx = αe⁻¹ + βe⁻¹/² + γ where α, β, γ are integers and [x] denotes the greatest integer less than or equal to x, then the value of α + β + γ is equal to:
If the integral ∫[0 to 10] (sin(2πx)) / (e^(x-[x])) dx = αe⁻¹ + βe⁻¹/² + γ where α, β, γ are integers and [x] denotes the greatest integer less than or equal to x, then the value of α + β + γ is equal to:
I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the in
Upvote 
If the equation a|z|² + α'z + αz' + d = 0 represents a circle where a, d are real constants, then which of the following condition is correct?
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If the equation a|z|² + α'z + αz' + d = 0 represents a circle where a, d are real constants, then which of the following condition is correct?
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The general equation of a circle is given by:
az z? + α? z + αz? + d = 0
This can be rewritten as:
z? + (α? /a)z + (α/a)z? + d/a = 0
From this, we can identify the centre and radius:
Centre = -α/a
Radius = √ (|-α/a|² - d/a)
For a real circle to exist, the term under the square root must be non-negative:
|-α
Upvote The area of the triangle vertices A(z), B(iz) and C(z + iz) is:
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The area of the triangle vertices A(z), B(iz) and C(z + iz) is:
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Kindly consider the following figure
Upvote The product of the roots of the equation 9x² – 18|x| + 5 = 0, is:
The product of the roots of the equation 9x² – 18|x| + 5 = 0, is:
∴ x² = |x|² = t let
9t² - 18t + 5 = 0
(3t - 1) (3t - 5) = 0
|x| = 1/3, 5/3
Product of roots = (1/3) (-1/3) (5/3) (-5/3) = 25/81
Upvote Let z = x + iy be a non-zero complex number such that z² = i|z|², where i = √-1, then z lies on the:
Let z = x + iy be a non-zero complex number such that z² = i|z|², where i = √-1, then z lies on the:
Assuming the equation is (x + iy)² = I (x² + y²)
x² - y² + 2ixy = I (x² + y²)
Compare real and imaginary parts
x² - y² = 0 ⇒ x = y or x = -y
2xy = x² + y²
If x=y, then 2x² = x² + x², which is true for all x.
If x=-y, then -2y² = y² + y² = 2y², which implies 4y²=0, so y=0 and x=0.
The non-trivial solution i
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