Revision: Complex Numbers and Quadratic Equations JEE Main Complex Numbers and Quadratic Equations

z = x + iy, x, y∈ R and \[i=\sqrt{-1}\] is called a complex number. 

x ⇒ Real Part Re(z)

iy ⇒ Imaginary Part Im(z)

If Re(z) = x = 0, then the complex number z is purely imaginary.

If Im(z) =y = 0, then complex number z is purely real.

Integral powers of iota (i):

\[\mathrm{i}^2=-1\]

\[\mathrm{i}^3=-\mathrm{i}\]

\[\mathrm{i}^{4}=1\]

In general,

\[1^{4n}=1\], \[\mathrm{i^{4n+1}=i}\], \[\mathrm{i^{4n+2}=-1}\], \[\mathrm{i^{4n+3}=-i}\] ...where n ∈ N

The cube roots of unity are the solutions of the equation
x³ = 1

They are: 1, \[\frac{-1+i\sqrt{3}}{2}\], \[\frac{-1-i\sqrt{3}}{2}\]

They are denoted by 1, ω, ω²

For the quadratic equation ax² + bx + c = 0, a ≠ 0; the expression b² − 4ac is called the discriminant and is, in general, denoted by the letter 'D'.

Thus, discriminant D = b² − 4ac.

The quadratic equation whose roots are α and β is

x² − (α+β)x + αβ = 0

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Given the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² – 4ac = 0

⇒ (r – p)² – 4 × (q – r) × (p – q) = 0

⇒ r² + p² – 2pr – 4[qp – q² – rp + qr] = 0

⇒ r² + p² – 2pr – 4qp + 4q² + 4rp – 4qr = 0

⇒ r² + p² + 2pr – 4qp – 4qr + 4q² = 0

⇒ (p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

⇒ y² – 4qy + 4q² = 0

⇒ (y – 2q)² = 0

⇒ y – 2q = 0

⇒ y = 2q

⇒ p + r = 2q

Hence proved.

Let √(a + ib) = x + iy

1. Square both sides
   (x + iy)² = a + ib
2. Expand
   x² − y² + 2ixy = a + ib
3. Equate real and imaginary parts
   x² − y² = a
   2xy = b
4. Solve these equations to find x and y
5. Then, √(a + ib) = ±(x + iy)

• ω³ = 1
• 1 + ω + ω² = 0
• ω² = 1/ω
• ω̄ = ω² and \[\left(\overline{\omega}\right)^2=\omega\]
• ω³ⁿ = 1
  ω³ⁿ⁺¹ = ω
  ω³ⁿ⁺² = ω²
• ω + ω² = −1
• ωω² = 1
• arg(ω) = \[\frac{2\pi}{3}\]
  arg(ω²) = \[\frac{4\pi}{3}\]

D = b² – 4ac 

1. Write the given equation in the standard form

   ax² + bx + c = 0

2. Identify the values of a, b, and c.

3. Find the value of the discriminant

   D = b² − 4ac

4. Substitute the values of a, b, and D in the formula

   \[x=\frac{-b-\sqrt{b^2-4ac}}{2a}\]​

5. Simplify to obtain the roots.