Quadratic Equations with Complex Numbers

Solving quadratic equations using complex numbers and the quadratic formula

Introduction to Quadratic Equations with Complex Numbers

One of the most important applications of complex numbers is solving quadratic equations. Complex numbers allow us to find solutions to equations that have no real solutions.

A fundamental example is the equation \(x^2 = -1\), which has no solution in the real numbers but has two complex solutions: \(x = i\) and \(x = -i\).

Key Insight:

Every quadratic equation with real coefficients has exactly two solutions in the complex numbers (counting multiplicity). These solutions may be real or complex conjugates.

Simple Quadratic Equations

Let us start with simple quadratic equations where the variable appears only as \(z^2\).

Example 1: \(z^2 = -1\)

Solving \(z^2 = -1\)

We need to find all complex numbers that, when squared, equal \(-1\).

\(\displaystyle z = \pm i\)

Verification:

  • \(i^2 = -1\) ✓
  • \((-i)^2 = (-1)^2 \cdot i^2 = 1 \cdot (-1) = -1\) ✓

Solutions: \(z_1 = i\) and \(z_2 = -i\) are conjugate complex numbers.

Example 2: \(z^2 = -2\)

Solving \(z^2 = -2\)

We rewrite the equation:

\(\displaystyle z^2 = -2 = 2 \cdot (-1)\)

Taking the square root:

\(\displaystyle z = \pm\sqrt{2} \cdot i = \pm i\sqrt{2}\)

Solutions: \(z_1 = i\sqrt{2}\) and \(z_2 = -i\sqrt{2}\)

Completing the Square Method

For more complex quadratic equations of the form \(az^2 + bz + c = 0\), we can use the method of completing the square to solve them.

Example: Solve \(z^2 - 6z + 13 = 0\)

1Complete the square
Identify the middle term: \(-6z = -2 \cdot 3 \cdot z\)
Working through the steps
\(\displaystyle z^2 - 6z + 13 = 0\)

Step 1: Rewrite with completing the square

\(\displaystyle z^2 - 6z + 9 - 9 + 13 = 0\)

Step 2: Factor the perfect square

\(\displaystyle (z - 3)^2 - 9 + 13 = 0\)

Step 3: Simplify

\(\displaystyle (z - 3)^2 + 4 = 0\)

Step 4: Isolate the squared term

\(\displaystyle (z - 3)^2 = -4 = 4 \cdot (-1)\)

Step 5: Take the square root

\(\displaystyle z - 3 = \pm 2i\)

Step 6: Solve for z

\(\displaystyle z = 3 \pm 2i\)

Solutions: \(z_1 = 3 + 2i\) and \(z_2 = 3 - 2i\)

The Quadratic Formula

For a general quadratic equation \(az^2 + bz + c = 0\) with \(a, b, c\) real and \(a \neq 0\), we can derive a general formula using completing the square.

The Quadratic Formula:

The solutions to \(az^2 + bz + c = 0\) are:

Quadratic Formula:
\(\displaystyle z_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

The Discriminant

The term under the square root is called the discriminant:

Discriminant:

\(\displaystyle \Delta = D = b^2 - 4ac\)

The discriminant determines the nature of the solutions:

\(D > 0\)
Two distinct real solutions

Both roots are real numbers
\(D = 0\)
One repeated real solution

\(z_1 = z_2 = -\frac{b}{2a}\)
\(D < 0\)
Two complex conjugate solutions

Solutions are complex conjugates

Quadratic Formula Examples

Example 1: Two Real Solutions (\(D > 0\))

Solve \(z^2 - 5z + 6 = 0\)

Identify coefficients: \(a = 1, b = -5, c = 6\)

Calculate discriminant:

\(\displaystyle D = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0\)

Apply formula:

\(\displaystyle z_{1,2} = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}\)

Solutions: \(z_1 = 3\) and \(z_2 = 2\)

Example 2: One Repeated Solution (\(D = 0\))

Solve \(z^2 - 4z + 4 = 0\)

Identify coefficients: \(a = 1, b = -4, c = 4\)

Calculate discriminant:

\(\displaystyle D = (-4)^2 - 4(1)(4) = 16 - 16 = 0\)

Apply formula:

\(\displaystyle z = \frac{4}{2} = 2\)

Solution: \(z_1 = z_2 = 2\) (double root)

Example 3: Complex Conjugate Solutions (\(D < 0\))

Solve \(z^2 - 4z + 5 = 0\)

Identify coefficients: \(a = 1, b = -4, c = 5\)

Calculate discriminant:

\(\displaystyle D = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0\)

Apply formula:

\(\displaystyle z_{1,2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i\)

Solutions: \(z_1 = 2 + i\) and \(z_2 = 2 - i\) (complex conjugates)

Quadratic Equation Summary

Discriminant Condition Nature of Solutions Example
\(\Delta = b^2 - 4ac\) \(\Delta > 0\) Two distinct real solutions \(z^2 - 5z + 6 = 0\) → \(z = 2, 3\)
\(\Delta = 0\) \(\Delta = 0\) One repeated real solution \(z^2 - 4z + 4 = 0\) → \(z = 2\)
\(\Delta < 0\) \(\Delta < 0\) Two complex conjugate solutions \(z^2 - 4z + 5 = 0\) → \(z = 2 \pm i\)

Applications and Key Insights

Why Complex Numbers Matter

Before complex numbers were introduced, quadratic equations with negative discriminants seemed unsolvable. With complex numbers, we can now:

  • Solve ALL quadratic equations with real coefficients
  • Find patterns in conjugate solutions
  • Apply quadratic equations in engineering and physics
  • Understand the fundamental theorem of algebra

Fundamental Theorem of Algebra

Important Theorem:

Every polynomial of degree \(n\) with complex coefficients has exactly \(n\) roots (counting multiplicity) in the complex numbers. This theorem is why complex numbers are so important in mathematics.

Practice Problems

Solve These Quadratic Equations
  • \(z^2 = 16\) → Answer: \(z = \pm 4\)
  • \(z^2 = -9\) → Answer: \(z = \pm 3i\)
  • \(z^2 + 2z + 2 = 0\) → Answer: \(z = -1 \pm i\)
  • \(z^2 - 2z + 10 = 0\) → Answer: \(z = 1 \pm 3i\)
  • \(2z^2 + 3z + 2 = 0\) → Answer: \(z = -\frac{1}{2}, -1\) (both real)
  • \(z^2 + 4 = 0\) → Answer: \(z = \pm 2i\)




























Is this page helpful?            
Thank you for your feedback!

Sorry about that

How can we improve it?