Introduction

A polynomial equation of degree two is called a quadratic equation. The general form is ax2+bx+c=0ax^2+bx+c=0, where a,b,ca, b, c are coefficients and a≠0a \ne 0. In the context of real numbers, this equation has no solution if the discriminant (D=b2−4acD = b^2-4ac) is negative. However, in the complex number system, a solution always exists.

The Quadratic Formula for Complex Roots

When the discriminant D=b2−4ac<0D = b^2-4ac < 0, the roots of the quadratic equation are a pair of complex conjugates. We can write D=−(4ac−b2)=i4ac−b2\sqrt{D} = \sqrt{-(4ac-b^2)} = i\sqrt{4ac-b^2}. The roots are then given by the familiar quadratic formula:

x=−b±b2−4ac2a=−b±i4ac−b22ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b \pm i\sqrt{4ac-b^2}}{2a}

The Fundamental Theorem of Algebra

This theorem states that a polynomial equation of degree n has n roots in the complex number system (counting roots with multiplicity).

  • For a quadratic equation (degree 2), there are always two roots.

Conjugate Root Theorem

If a quadratic equation has real coefficients (a,b,c∈Ra, b, c \in \mathbb{R}), then its complex roots always occur in conjugate pairs. If p+iqp+iq is a root, then p−iqp-iq must also be a root.

Example 1: Basic Equation with Complex Roots

Question: Solve the quadratic equation x2+x+1=0x^2 + x + 1 = 0.

Solution: Here, a=1,b=1,c=1a=1, b=1, c=1.

The discriminant is D=b2−4ac=12−4(1)(1)=−3D = b^2-4ac = 1^2 - 4(1)(1) = -3.

Using the quadratic formula:

x=−1±−32(1)=−1±i32x = \frac{-1 \pm \sqrt{-3}}{2(1)} = \frac{-1 \pm i\sqrt{3}}{2}

The two roots are −12+i32-\frac{1}{2} + i\frac{\sqrt{3}}{2} and −12−i32-\frac{1}{2} - i\frac{\sqrt{3}}{2}. (These are the complex cube roots of unity, ω\omega and ω2\omega^2)

Example 2: Forming an Equation from a Root

Question: Find the quadratic equation with real coefficients which has 3+2i3+2i as one of its roots.

Solution: Since the coefficients are real, the complex roots must occur in conjugate pairs. If one root is z1=3+2iz_1 = 3+2i, the other root must be z2=z1‾=3−2iz_2 = \overline{z_1} = 3-2i.

The quadratic equation can be written as (x−z1)(x−z2)=0(x-z_1)(x-z_2)=0, or more easily using the sum and product of roots:

x2−(Sum of roots)x+(Product of roots)=0x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0

  • Sum: z1+z2=(3+2i)+(3−2i)=6z_1+z_2 = (3+2i) + (3-2i) = 6.
  • Product: z1z2=(3+2i)(3−2i)=32−(2i)2=9−4i2=9+4=13z_1z_2 = (3+2i)(3-2i) = 3^2 - (2i)^2 = 9 - 4i^2 = 9+4=13.

The equation is x2−6x+13=0x^2 - 6x + 13 = 0.

Example 3: Equation with Complex Coefficients

Question: Solve the equation z2−(3+i)z+(2+2i)=0z^2 - (3+i)z + (2+2i) = 0.

Solution: We use the quadratic formula where a=1,b=−(3+i),c=2+2ia=1, b=-(3+i), c=2+2i.

D=b2−4ac=(−(3+i))2−4(1)(2+2i)D = b^2-4ac = (-(3+i))^2 - 4(1)(2+2i)

=(9+6i+i2)−8−8i=(8+6i)−8−8i=−2i= (9+6i+i^2) - 8 - 8i = (8+6i) - 8 - 8i = -2i

Now we need to find −2i\sqrt{-2i}. Let −2i=x+iy\sqrt{-2i} = x+iy.

Then −2i=(x2−y2)+2ixy-2i = (x^2-y^2)+2ixy. So, x2−y2=0x^2-y^2=0 and 2xy=−2  ⟹  xy=−12xy=-2 \implies xy=-1.

From the first part, x=±yx=\pm y. Since their product is negative, they must have opposite signs.

Thus, x=−yx=-y. Substituting this into xy=−1xy=-1 gives −y2=−1  ⟹  y=±1-y^2=-1 \implies y=\pm 1. The roots are (x=1,y=−1)(x=1, y=-1) and (x=−1,y=1)(x=-1, y=1). So −2i=±(1−i)\sqrt{-2i} = \pm(1-i).

Now, we find the roots of the equation:

z=(3+i)±(1−i)2z = \frac{(3+i) \pm (1-i)}{2}

  • Root 1: z1=(3+i)+(1−i)2=42=2z_1 = \frac{(3+i) + (1-i)}{2} = \frac{4}{2} = 2.
  • Root 2: z2=(3+i)−(1−i)2=2+2i2=1+iz_2 = \frac{(3+i) - (1-i)}{2} = \frac{2+2i}{2} = 1+i.