Complex Numbers and Quadratic Equations – CBSE NCERT Study Resources

We studied that for a quadratic equation ax² + bx + c = 0, the sum and product of roots can be derived using Vieta's formulas. If α and β are the roots, then α + β = -b/a and αβ = c/a.

Let’s verify this for 2x² - 5x + 3 = 0. Here, a = 2, b = -5, c = 3. Sum of roots = -(-5)/2 = 5/2. Product of roots = 3/2. Solving the equation, roots are 1 and 1.5. Sum = 2.5 (5/2), product = 1.5 (3/2), matching Vieta's formulas.

This proof is fundamental in algebra, linking coefficients to roots. It simplifies solving quadratic equations without factorization.

We learned that for complex numbers z₁ = a + bi and z₂ = c + di, the product z₁ × z₂ = (ac - bd) + (ad + bc)i. Modulus is √(Re² + Im²), and argument is tan⁻¹(Im/Re).

For z₁ = 3 + 4i and z₂ = 1 - 2i, product = (3×1 - 4×(-2)) + (3×(-2) + 4×1)i = 11 - 2i. Modulus = √(11² + (-2)²) = √125 ≈ 11.18. Argument = tan⁻¹(-2/11) ≈ -0.18 radians. [Diagram: Vector from origin to (11, -2)].

This shows how complex multiplication scales magnitudes and adds angles, crucial in polar form applications like signal processing.

We studied that a complex number is of the form z = a + ib, where a, b ∈ ℝ. Let two complex numbers be z₁ = a + ib and z₂ = c + id.

• Sum: z₁ + z₂ = (a + c) + i(b + d), which is clearly a complex number.
• Product: z₁ × z₂ = (ac - bd) + i(ad + bc), also a complex number.

Our textbook shows that the set of complex numbers is closed under addition and multiplication, confirming the proof.

The discriminant D = b² - 4ac determines the nature of roots. For 2x² - 4x + 3 = 0, a = 2, b = -4, c = 3.

• Calculation: D = (-4)² - 4(2)(3) = 16 - 24 = -8.
• Roots: Since D < 0, the roots are complex conjugates.

We confirm that the equation has no real roots, aligning with the graphical representation of a parabola not intersecting the x-axis.

We know z = a + ib can be represented geometrically with modulus r = √(a² + b²) and argument θ = tan⁻¹(b/a).

• Conversion: Using Euler's formula, z = r(cosθ + isinθ).
• Polar Form: Thus, z = re^(iθ), where r is modulus and θ is argument.

Our textbook shows this form simplifies multiplication and division of complex numbers, highlighting its practical utility.

We studied that for a quadratic equation ax² + bx + c = 0, the sum and product of roots can be derived using Vieta's formulas. Let the roots be α and β.

• Sum of roots: α + β = -b/a (coefficient of x with sign change divided by coefficient of x²).
• Product of roots: αβ = c/a (constant term divided by coefficient of x²).
• For x² - 5x + 6 = 0, sum = 5, product = 6. Roots are 2 and 3 by factorization.

This proof is fundamental in algebra and simplifies solving quadratic equations without factorization.

Our textbook shows that complex roots occur in conjugate pairs for real coefficients. The equation z² + 4z + 5 = 0 can be solved using the quadratic formula.

• Roots: z = [-4 ± √(16 - 20)]/2 = -2 ± i.
• On the Argand plane, roots are at (-2, 1) and (-2, -1), symmetric about the real axis.

This demonstrates the fundamental theorem of algebra and the geometric symmetry of complex conjugates, essential for advanced topics like signal processing.

We learned that a quadratic equation has equal roots if the discriminant D = b² - 4ac = 0. This ensures a perfect square trinomial.

• For x² + kx + 9 = 0, D = k² - 36 = 0 → k = ±6.
• Equal roots imply the parabola touches the x-axis at one point (vertex).

This condition is vital in optimization problems and physics (e.g., projectile motion), where double roots indicate critical points.

We studied that if a complex number is z = a + ib, its conjugate is z̄ = a - ib. The sum (z + z̄) and product (z × z̄) of a complex number and its conjugate must yield real numbers.

• Sum: (a + ib) + (a - ib) = 2a (real)
• Product: (a + ib)(a - ib) = a² + b² (real)

For z = 3 + 4i and z̄ = 3 - 4i:

• Sum: 6 (real)
• Product: 9 + 16 = 25 (real)

The quadratic formula x = [-b ± √(b² - 4ac)]/2a gives roots of ax² + bx + c = 0. The discriminant D = b² - 4ac determines the nature of roots.

• For 2x² - 5x + 3 = 0, a=2, b=-5, c=3.
• D = 25 - 24 = 1 > 0 ⇒ real, distinct roots.
• Roots: x = [5 ± 1]/4 ⇒ x = 1.5, 1.

Since D > 0, our textbook confirms two real roots. The calculation matches the theoretical prediction.

Polar form of z = a + ib is r(cosθ + isinθ), where r = √(a² + b²) and θ = tan⁻¹(b/a). De Moivre's Theorem states zⁿ = rⁿ(cos(nθ) + isin(nθ)).

• For z = 1 + √3i, r = 2, θ = π/3.
• Polar form: 2(cos(π/3) + isin(π/3)).
• Square: z² = 4(cos(2π/3) + isin(2π/3)) = -2 + 2√3i.

Direct multiplication (1 + √3i)² also gives -2 + 2√3i, validating De Moivre's Theorem.

To prove that the sum and product of two complex numbers are also complex numbers, let us consider z₁ = a + ib and z₂ = c + id, where a, b, c, d are real numbers and i is the imaginary unit (i² = -1).

Sum of two complex numbers:
z₁ + z₂ = (a + ib) + (c + id) = (a + c) + i(b + d).
Since a + c and b + d are real numbers, the sum is also a complex number.

Product of two complex numbers:
z₁ × z₂ = (a + ib)(c + id) = ac + iad + ibc + i²bd = (ac - bd) + i(ad + bc).
Since ac - bd and ad + bc are real numbers, the product is also a complex number.

Verification with an example:
Let z₁ = 3 + 4i and z₂ = 1 - 2i.
Sum: (3 + 1) + i(4 - 2) = 4 + 2i (a complex number).
Product: (3)(1) + (3)(-2i) + (4i)(1) + (4i)(-2i) = 3 - 6i + 4i - 8i² = 3 - 2i + 8 = 11 - 2i (a complex number).

The given quadratic equation is 2x² - 5x + 3 = 0. To solve it using the quadratic formula, we follow these steps:

Step 1: Identify coefficients
Compare with the standard form ax² + bx + c = 0.
Here, a = 2, b = -5, and c = 3.

Step 2: Apply the quadratic formula
The quadratic formula is:
x = [-b ± √(b² - 4ac)] / (2a).
Substitute the values:
x = [5 ± √((-5)² - 4 × 2 × 3)] / (4).
x = [5 ± √(25 - 24)] / 4.
x = [5 ± √1] / 4.

Step 3: Calculate roots
For the positive root:
x = (5 + 1) / 4 = 6 / 4 = 3/2.
For the negative root:
x = (5 - 1) / 4 = 4 / 4 = 1.

Verification by substitution:
For x = 3/2:
2(3/2)² - 5(3/2) + 3 = 2(9/4) - 15/2 + 3 = 9/2 - 15/2 + 3 = -6/2 + 3 = -3 + 3 = 0.
For x = 1:
2(1)² - 5(1) + 3 = 2 - 5 + 3 = 0.
Both roots satisfy the equation, confirming their correctness.

In the Argand plane, a complex number z = a + ib is represented as a point with coordinates (a, b), where the x-axis represents the real part and the y-axis represents the imaginary part.

Example: Plotting z = 3 + 4i
The complex number z = 3 + 4i corresponds to the point (3, 4) in the Argand plane.

Modulus (|z|):
The modulus is the distance from the origin to the point (3, 4).
|z| = √(a² + b²) = √(3² + 4²) = √(9 + 16) = √25 = 5.

Argument (θ):
The argument is the angle between the positive real axis and the line joining the origin to (3, 4).
θ = tan^-1(b/a) = tan^-1(4/3).
Since z lies in the first quadrant, θ is the principal value (≈ 53.13°).

Geometric Interpretation:
The modulus represents the magnitude of the complex number, while the argument represents its direction in the plane.

To prove that the sum and product of two complex numbers are also complex numbers, let us consider z₁ = a + ib and z₂ = c + id, where a, b, c, and d are real numbers.

Sum of two complex numbers:
z₁ + z₂ = (a + ib) + (c + id)
= (a + c) + i(b + d)
Since a + c and b + d are real numbers, the sum is a complex number.

Product of two complex numbers:
z₁ × z₂ = (a + ib)(c + id)
= ac + i(ad) + i(bc) + i²(bd)
= (ac - bd) + i(ad + bc) (since i² = -1)
Since ac - bd and ad + bc are real numbers, the product is a complex number.

Verification for z₁ = 3 + 4i and z₂ = 1 - 2i:

• Sum: (3 + 1) + i(4 - 2) = 4 + 2i (a complex number)
• Product: (3 × 1 - 4 × (-2)) + i(3 × (-2) + 4 × 1) = (3 + 8) + i(-6 + 4) = 11 - 2i (a complex number)

To solve the quadratic equation x² - 4x + 5 = 0, we use the quadratic formula:
x = [ -b ± √(b² - 4ac) ] / 2a
Here, a = 1, b = -4, and c = 5.

Step 1: Calculate the discriminant (D):
D = b² - 4ac
= (-4)² - 4(1)(5)
= 16 - 20
= -4

Step 2: Find the roots using the quadratic formula:
Since D < 0, the roots are complex.
x = [ -(-4) ± √(-4) ] / 2(1)
= [ 4 ± 2i ] / 2 (since √(-4) = 2i)
= 2 ± i

Roots: x = 2 + i and x = 2 - i

Representation on the complex plane:
The roots are plotted as points where:

• 2 + i corresponds to (2, 1)
• 2 - i corresponds to (2, -1)

To prove that the square of any complex number a + ib is also a complex number, we follow these steps:

To solve the quadratic equation 2x² - 5x + 3 = 0 using the quadratic formula, follow these steps:

To solve the quadratic equation 2x² + 3x + 1 = 0, we use the quadratic formula:

x = [-b ± √(b² - 4ac)] / (2a)

Here, a = 2, b = 3, and c = 1.

Step 1: Calculate the discriminant (D):
D = b² - 4ac = (3)² - 4(2)(1) = 9 - 8 = 1

Step 2: Substitute the values into the quadratic formula:
x = [-3 ± √1] / (4)

Step 3: Simplify the roots:
x₁ = (-3 + 1)/4 = -2/4 = -1/2
x₂ = (-3 - 1)/4 = -4/4 = -1

Verification:

For x = -1/2:
2(-1/2)² + 3(-1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = 0

For x = -1:
2(-1)² + 3(-1) + 1 = 2(1) - 3 + 1 = 2 - 3 + 1 = 0

Both roots satisfy the original equation, confirming their correctness.

To solve the quadratic equation 2x² + 3x + 1 = 0, we use the quadratic formula:

x = [-b ± √(b² - 4ac)] / (2a)

Here, a = 2, b = 3, and c = 1.

Step 1: Calculate the discriminant (D)
D = b² - 4ac = (3)² - 4(2)(1) = 9 - 8 = 1

Step 2: Apply the quadratic formula
x = [-3 ± √1] / (4) = [-3 ± 1] / 4

This gives two solutions:
x₁ = (-3 + 1)/4 = -2/4 = -1/2
x₂ = (-3 - 1)/4 = -4/4 = -1

Verification:

For x = -1/2:
2(-1/2)² + 3(-1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = 0

For x = -1:
2(-1)² + 3(-1) + 1 = 2(1) - 3 + 1 = 2 - 3 + 1 = 0

Both roots satisfy the original equation, confirming their correctness.

To solve the quadratic equation 2x² + 3x + 1 = 0, we use the quadratic formula:

x = [-b ± √(b² - 4ac)] / (2a)

Here, a = 2, b = 3, and c = 1.

Step 1: Calculate the discriminant (D):

D = b² - 4ac = (3)² - 4(2)(1) = 9 - 8 = 1

Since D > 0, the equation has two distinct real roots.

Step 2: Apply the quadratic formula:

x = [-3 ± √1] / (4)

This gives two solutions:

x₁ = (-3 + 1)/4 = -2/4 = -0.5

x₂ = (-3 - 1)/4 = -4/4 = -1

Verification of roots:

For x₁ = -0.5:

2(-0.5)² + 3(-0.5) + 1 = 2(0.25) - 1.5 + 1 = 0.5 - 1.5 + 1 = 0

For x₂ = -1:

2(-1)² + 3(-1) + 1 = 2(1) - 3 + 1 = 2 - 3 + 1 = 0

Both roots satisfy the original equation, confirming their correctness.

Additional Insight: The discriminant (D) not only determines the nature of the roots but also helps in understanding whether the quadratic can be factored easily. Here, D = 1 (a perfect square) implies the roots are rational and the equation can be factored as (2x + 1)(x + 1) = 0.