Section 3.1: The Complex Number System – Precalculus Study Notes

Section 3.1: The Complex Number System

Objectives

This section introduces the complex-number system and demonstrates how to perform computations involving complex numbers, including addition, subtraction, multiplication, and simplification of powers of i.

The Complex-Number System

Some algebraic functions have zeros that are not real numbers. The complex-number system is used to find zeros of functions that do not have real-number solutions. If the graph of a function does not cross the x-axis, it has no x-intercepts and thus no real-number zeros.

• Complex numbers extend the real number system to include solutions to equations like $x^2 + 1 = 0$.

• The imaginary unit i is defined as $i = \sqrt{-1}$.

Expressing Square Roots of Negative Numbers in Terms of i

Any square root of a negative number can be written in terms of i:

• $\sqrt{-a} = i\sqrt{a}$, where $a > 0$

Examples:

• $\sqrt{-7} = i\sqrt{7}$ or $\sqrt{7}i$

• $\sqrt{-16} = i\cdot 4 = 4i$

• $-\sqrt{-13} = -i\sqrt{13}$ or $-\sqrt{13}i$

• $-\sqrt{-64} = -i\cdot 8 = -8i$

• $\sqrt{-48} = i\sqrt{16 \cdot 3} = i\cdot 4\sqrt{3} = 4i\sqrt{3}$ or $4\sqrt{3}i$

Complex Numbers: Definition and Classification

A complex number is a number of the form $a + bi$, where $a$ and $b$ are real numbers. The number $a$ is the real part and $b$ is the imaginary part.

Type	Form	Conditions
Imaginary Number	$a + bi$	$a \neq 0$, $b \neq 0$
Pure Imaginary Number	$a + bi$	$a = 0$, $b \neq 0$

Addition and Subtraction of Complex Numbers

Complex numbers obey the commutative, associative, and distributive laws. To add or subtract complex numbers, combine the real parts and the imaginary parts separately, just as with binomials.

• $(a + bi) + (c + di) = (a + c) + (b + d)i$

• $(a + bi) - (c + di) = (a - c) + (b - d)i$

Examples:

• $(8 + 6i) + (3 + 2i) = (8 + 3) + (6 + 2)i = 11 + 8i$

• $(4 + 5i) - (6 - 3i) = (4 - 6) + [5i - (-3i)] = -2 + 8i$

Multiplication of Complex Numbers

When multiplying square roots of real numbers, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, but this does not hold for negative radicands. For complex numbers, use $i^2 = -1$.

• $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Examples:

• $\sqrt{-16} \cdot \sqrt{-25} = (4i) \cdot (5i) = 20i^2 = -20$

• $(1 + 2i)(1 + 3i) = 1 + 3i + 2i + 6i^2 = 1 + 5i - 6 = -5 + 5i$

• $(3 - 7i)^2 = 9 - 42i + 49i^2 = 9 - 42i - 49 = -40 - 42i$

Simplifying Powers of i

Recall that $i = \sqrt{-1}$ and $i^2 = -1$. Powers of $i$ cycle every four terms:

• $i^1 = i$

• $i^2 = -1$

• $i^3 = -i$

• $i^4 = 1$

• $i^5 = i$ (and so on)

Examples:

• $i^{37} = i^{36} \cdot i = (i^4)^9 \cdot i = 1^9 \cdot i = i$

• $i^{58} = (i^2)^{29} = (-1)^{29} = -1$

Complex Conjugates

The conjugate of a complex number $a + bi$ is $a - bi$. The product of a complex number and its conjugate is always a real number:

• $(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2$

Examples:

• $(5 + 7i)(5 - 7i) = 25 + 49 = 74$

• $(8i)(-8i) = -64i^2 = 64$

Division of Complex Numbers

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator to rationalize it.

Example:

• Divide $2 - 5i$ by $1 - 6i$:

$\frac{2 - 5i}{1 - 6i} \cdot \frac{1 + 6i}{1 + 6i} = \frac{(2 - 5i)(1 + 6i)}{(1 - 6i)(1 + 6i)}$

Numerator: $(2 - 5i)(1 + 6i) = 2 + 12i - 5i - 30i^2 = 2 + 7i + 30 = 32 + 7i$ Denominator: $(1 - 6i)(1 + 6i) = 1 - 36i^2 = 1 + 36 = 37$ Final answer: $\frac{32 + 7i}{37}$

*Additional info: The notes cover all foundational operations with complex numbers, including their definition, arithmetic, powers, and conjugates, which are essential for Precalculus students studying polynomial equations and functions.*