Complex Numbers and Quadratic Equations: Operations and Standard Forms

Complex Numbers

Operations with Complex Numbers

Complex numbers are numbers of the form a + bi, where a and b are real numbers, and i is the imaginary unit defined by $i^2 = -1$. Operations with complex numbers include addition, subtraction, multiplication, and division, and results are typically written in standard a + bi form.

• Addition/Subtraction: Combine real parts and imaginary parts separately.

• Multiplication: Use distributive property and the fact that $i^2 = -1$.

• Division: Multiply numerator and denominator by the conjugate of the denominator to rationalize.

Example: Perform the indicated operation and write the result in standard a + bi form.

• Example 1: $(7 + 4i) - (2 - 3i)$ $= (7 - 2) + (4i - (-3i))$ $= 5 + 7i$

• Example 2: $5\sqrt{-8} - 2\sqrt{-36}$ Recall $\sqrt{-a} = i\sqrt{a}$ for $a > 0$. $5\sqrt{-8} = 5i\sqrt{8} = 5i \times 2\sqrt{2} = 10i\sqrt{2}$ $-2\sqrt{-36} = -2i\sqrt{36} = -2i \times 6 = -12i$ $10i\sqrt{2} - 12i$ (cannot combine further; this is in a + bi form with $a = 0$)

Standard Form: Always express complex numbers as $a + bi$.

Quadratic Equations

Solving Quadratic Equations with Complex Solutions

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$. When the discriminant $b^2 - 4ac$ is negative, the solutions are complex numbers.

• Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Complex Solutions: If $b^2 - 4ac < 0$, then $\sqrt{b^2 - 4ac}$ is imaginary, and solutions are complex conjugates.

Example: Solve $x^2 + 4x + 13 = 0$.

• Discriminant: $4^2 - 4 \times 1 \times 13 = 16 - 52 = -36$

• Solutions: $x = \frac{-4 \pm \sqrt{-36}}{2}$ $= \frac{-4 \pm 6i}{2}$ $= -2 \pm 3i$

Summary Table: Complex Number Operations

Operation	Formula	Example
Addition	$(a + bi) + (c + di) = (a + c) + (b + d)i$	$(3 + 2i) + (1 + 4i) = 4 + 6i$
Subtraction	$(a + bi) - (c + di) = (a - c) + (b - d)i$	$(5 + 7i) - (2 + 3i) = 3 + 4i$
Multiplication	$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$	$(2 + 3i)(4 + i) = (2\times4 - 3\times1) + (2\times1 + 3\times4)i = 5 + 14i$
Division	$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)}$	$\frac{1 + 2i}{3 - i} = \frac{(1 + 2i)(3 + i)}{(3 - i)(3 + i)} = \frac{3 + i + 6i + 2i^2}{9 + 1} = \frac{3 + 7i - 2}{10} = \frac{1 + 7i}{10}$

Key Concepts

• Imaginary Unit: $i$ is defined such that $i^2 = -1$.

• Complex Conjugate: For $a + bi$, the conjugate is $a - bi$.

• Standard Form: Always write complex numbers as $a + bi$.

• Quadratic Equations: Use the quadratic formula to find complex solutions when the discriminant is negative.

Additional info: Some context and examples have been inferred based on standard Precalculus curriculum and the fragmentary nature of the provided material.