MAT 142: Complex Numbers

Reminders

At our last meeting, we discussed each of the following:

Linear functions
- The slope of a line
- The \(y\)-intercept of a line
- Writing the equation of a line in slope-intercept form
- Writing the equation of a line in point-slope form
Connecting linear functions to the graphical transformation framework
Parallel and perpendicular lines
Solving linear equations and the corresponding geometric interpretation

Try the following warm-up problems.

Problem 1: Simplify the expression \(\left(2 + 3x\right) + \left(7 - 2x\right)\).

Problem 2: Simplify the expression \(\left(2 + 3x\right)\left(7 - 2x\right)\).

Note: These warm-up problems are designed to preview today’s topic. Keep your answers in mind as we move forward.

Objectives

Today we take a brief but important detour to introduce the complex numbers before moving into our study of polynomial functions.

After today’s class meeting, you should be able to:

Write and recognize complex numbers of the form \(a + bi\), where \(a\), \(b\) are real numbers and \(i = \sqrt{\left(-1\right)}\).
Identify and construct the complex conjugate of a complex number.
Add, subtract, multiply, and divide complex numbers, expressing all results in the form \(a + bi\).

What Are Complex Numbers?

Definition (Complex Number): A complex number is a number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i = \sqrt{\left(-1\right)}\).

The collection of all complex numbers is denoted \(\mathbb{C}\).

\(a\) is called the real part and \(b\) is called the imaginary part.
Real numbers are complex numbers with \(b = 0\). So \(\mathbb{R}\) is a subset of \(\mathbb{C}\) (or, in symbols, \(\mathbb{R} \subset \mathbb{C}\)).
Pure imaginary numbers have \(a = 0\), such as \(3i\) or \(-7i\).

Powers of \(i\):

\[i^0 = 1 \qquad i^1 = i \qquad i^2 = -1 \qquad i^3 = -i \qquad i^4 = 1 \qquad i^5 = i \qquad \cdots\]

The pattern repeats with period 4. This is the key fact that makes complex arithmetic work.

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) \end{align}\]

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) &= 5 + \left(-7i + 3i\right) \end{align}\]

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) &= 5 + \left(-7i + 3i\right)\\ &= \boxed{~5 - 4i~} \end{align}\]

Example 2: Compute the difference \(\left(1 + 18i\right) - \left(-3 - i\right)\).

Solution.

\[\begin{align} \left(1 + 18i\right) - \left(-3 - i\right) \end{align}\]

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) &= 5 + \left(-7i + 3i\right)\\ &= \boxed{~5 - 4i~} \end{align}\]

Example 2: Compute the difference \(\left(1 + 18i\right) - \left(-3 - i\right)\).

Solution.

\[\begin{align} \left(1 + 18i\right) - \left(-3 - i\right) &= 1 + 18i + 3 + i \end{align}\]

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) &= 5 + \left(-7i + 3i\right)\\ &= \boxed{~5 - 4i~} \end{align}\]

Example 2: Compute the difference \(\left(1 + 18i\right) - \left(-3 - i\right)\).

Solution.

\[\begin{align} \left(1 + 18i\right) - \left(-3 - i\right) &= 1 + 18i + 3 + i\\ &= \left(1 + 3\right) + \left(18i + i\right) \end{align}\]

Adding and Subtracting Complex Numbers

Strategy: Adding and subtracting complex numbers works just like combining like terms with a variable. Treat \(i\) as if it were a variable such as \(x\).

Combine real parts with real parts, and imaginary parts with imaginary parts.

Example 1: Compute the sum \(\left(5 - 7i\right) + \left(3i\right)\).

Solution. Combine like terms:

\[\begin{align} \left(5 - 7i\right) + \left(3i\right) &= 5 + \left(-7i + 3i\right)\\ &= \boxed{~5 - 4i~} \end{align}\]

Example 2: Compute the difference \(\left(1 + 18i\right) - \left(-3 - i\right)\).

Solution.

\[\begin{align} \left(1 + 18i\right) - \left(-3 - i\right) &= 1 + 18i + 3 + i\\ &= \left(1 + 3\right) + \left(18i + i\right)\\ &= \boxed{~4 + 19i~} \end{align}\]

Adding and Subtracting Practice

Try It! 1: Simplify \(\left(3 + 5i\right) + \left(2 - 9i\right)\)

Try It! 2: Simplify \(\left(4 - i\right) - \left(-1 + 6i\right)\)

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2 \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right) \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6 \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6\\ &= \boxed{~-18 - 26i~} \end{align}\]

Example 4: Simplify \(\left(5 - 8i\right)\left(5 + 8i\right)\).

Solution.

\[\begin{align} \left(5 - 8i\right)\left(5 + 8i\right) \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6\\ &= \boxed{~-18 - 26i~} \end{align}\]

Example 4: Simplify \(\left(5 - 8i\right)\left(5 + 8i\right)\).

Solution.

\[\begin{align} \left(5 - 8i\right)\left(5 + 8i\right) &= 25 + 40i - 40i - 64i^2 \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6\\ &= \boxed{~-18 - 26i~} \end{align}\]

Example 4: Simplify \(\left(5 - 8i\right)\left(5 + 8i\right)\).

Solution.

\[\begin{align} \left(5 - 8i\right)\left(5 + 8i\right) &= 25 + 40i - 40i - 64i^2\\ &= 25 -64\left(-1\right) \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6\\ &= \boxed{~-18 - 26i~} \end{align}\]

Example 4: Simplify \(\left(5 - 8i\right)\left(5 + 8i\right)\).

Solution.

\[\begin{align} \left(5 - 8i\right)\left(5 + 8i\right) &= 25 + 40i - 40i - 64i^2\\ &= 25 -64\left(-1\right)\\ &= 25 + 64 \end{align}\]

Multiplying Complex Numbers

Strategy: Multiply complex numbers just like polynomial expressions – distribute (multiply everything in one factor by everything in the other factor), then simplify using the fact that \(i^2 = -1\) to eliminate any \(i^2\) terms and express the result in the form \(a + bi\).

Example 3: Simplify \(\left(6 + 2i\right)\left(-4 - 3i\right)\).

Solution.

\[\begin{align} \left(6 + 2i\right)\left(-4 - 3i\right) &= -24 - 18i - 8i - 6i^2\\ &= -24 -26i - 6\left(-1\right)\\ &= -24 -26i + 6\\ &= \boxed{~-18 - 26i~} \end{align}\]

Example 4: Simplify \(\left(5 - 8i\right)\left(5 + 8i\right)\).

Solution.

\[\begin{align} \left(5 - 8i\right)\left(5 + 8i\right) &= 25 + 40i - 40i - 64i^2\\ &= 25 -64\left(-1\right)\\ &= 25 + 64\\ &= \boxed{~89~} \end{align}\]

The result is a real number. What was special about the two complex numbers we multiplied?

Multiplication Practice

Try It! 3: Simplify \(\left(2 + 7i\right)\left(3 - 4i\right)\)

Try It! 4: Simplify \(\left(1 - 3i\right)\left(1 + 3i\right)\)

Try It! 5: Simplify \(\left(9 - 2i\right)\left(9 + 2i\right)\)

Complex Conjugates

Definition (Complex Conjugate): The complex conjugate of the complex number \(a + bi\) is the complex number \(a - bi\).

In Example 4, the numbers \(5 - 8i\) and \(5 + 8i\) are complex conjugates of one another. Their product was the real number \(89\). This happened in Try It! 5 for the same reason.

Products of Complex Conjugates

The product of a complex number and its conjugate is always a real number.

\[\begin{align} \left(a + bi\right)\left(a - bi\right) \end{align}\]

Complex Conjugates

Definition (Complex Conjugate): The complex conjugate of the complex number \(a + bi\) is the complex number \(a - bi\).

In Example 4, the numbers \(5 - 8i\) and \(5 + 8i\) are complex conjugates of one another. Their product was the real number \(89\).

Products of Complex Conjugates

The product of a complex number and its conjugate is always a real number.

\[\begin{align} \left(a + bi\right)\left(a - bi\right) &= a^2 - abi + abi - b^2i^2 \end{align}\]

Complex Conjugates

Definition (Complex Conjugate): The complex conjugate of the complex number \(a + bi\) is the complex number \(a - bi\).

In Example 4, the numbers \(5 - 8i\) and \(5 + 8i\) are complex conjugates of one another. Their product was the real number \(89\).

Products of Complex Conjugates

The product of a complex number and its conjugate is always a real number.

\[\begin{align} \left(a + bi\right)\left(a - bi\right) &= a^2 - abi + abi - b^2i^2\\ &= a^2 -b^2\left(-1\right) \end{align}\]

Complex Conjugates

Definition (Complex Conjugate): The complex conjugate of the complex number \(a + bi\) is the complex number \(a - bi\).

In Example 4, the numbers \(5 - 8i\) and \(5 + 8i\) are complex conjugates of one another. Their product was the real number \(89\).

Products of Complex Conjugates

The product of a complex number and its conjugate is always a real number.

\[\begin{align} \left(a + bi\right)\left(a - bi\right) &= a^2 - abi + abi - b^2i^2\\ &= a^2 -b^2\left(-1\right)\\ &= \boxed{~a^2 + b^2~} \end{align}\]

This fact is what makes division of complex numbers possible.

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right) \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right)\\ &= \frac{\left(5 - 2i\right)\left(1 - i\right)}{\left(1 + i\right)\left(1 - i\right)} \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right)\\ &= \frac{\left(5 - 2i\right)\left(1 - i\right)}{\left(1 + i\right)\left(1 - i\right)}\\ &= \frac{5 - 5i - 2i + 2i^2}{1 -i + i -i^2} \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right)\\ &= \frac{\left(5 - 2i\right)\left(1 - i\right)}{\left(1 + i\right)\left(1 - i\right)}\\ &= \frac{5 - 5i -2i + 2i^2}{1 -i + i -i^2}\\ &= \frac{5 - 7i + 2\left(-1\right)}{1 - \left(-1\right)} \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right)\\ &= \frac{\left(5 - 2i\right)\left(1 - i\right)}{\left(1 + i\right)\left(1 - i\right)}\\ &= \frac{5 - 5i -2i + 2i^2}{1 -i + i -i^2}\\ &= \frac{5 - 7i + 2\left(-1\right)}{1 - \left(-1\right)}\\ &= \frac{3 - 7i}{2} \end{align}\]

Division

Strategy: To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real, allowing us to write the result in the form \(a + bi\).

Example 5: Simplify \(\displaystyle{\frac{5 - 2i}{1 + i}}\).

Solution. Multiply numerator and denominator by the conjugate of the denominator, \(1 - i\).

\[\begin{align} \frac{5 - 2i}{1 + i} &= \left(\frac{5 - 2i}{1 + i}\right) \cdot \left(\frac{1 - i}{1 - i}\right)\\ &= \frac{\left(5 - 2i\right)\left(1 - i\right)}{\left(1 + i\right)\left(1 - i\right)}\\ &= \frac{5 - 5i -2i + 2i^2}{1 -i + i -i^2}\\ &= \frac{5 - 7i + 2\left(-1\right)}{1 - \left(-1\right)}\\ &= \frac{3 - 7i}{2}\\ &= \boxed{~\frac{3}{2} - \frac{7}{2}i~} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right) \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)}\\ &= \frac{6i + 15i^2}{-9i^2} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)}\\ &= \frac{6i + 15i^2}{-9i^2}\\ &= \frac{6i + 15\left(-1\right)}{-9\left(-1\right)} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)}\\ &= \frac{6i + 15i^2}{-9i^2}\\ &= \frac{6i + 15\left(-1\right)}{-9\left(-1\right)}\\ &= \frac{6i - 15}{9} \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)}\\ &= \frac{6i + 15i^2}{-9i^2}\\ &= \frac{6i + 15\left(-1\right)}{-9\left(-1\right)}\\ &= \frac{6i - 15}{9}\\ &= \frac{-15}{9} + \frac{6}{9}i \end{align}\]

Division

Example 6: Simplify \(\displaystyle{\frac{-2 - 5i}{3i}}\).

Solution. Note that \(3i = 0 + 3i\), so its complex conjugate is \(0 - 3i = -3i\).

\[\begin{align} \frac{-2 - 5i}{3i} &= \left(\frac{-2 - 5i}{3i}\right) \cdot \left(\frac{-3i}{-3i}\right)\\ &= \frac{\left(-2 - 5i\right)\left(-3i\right)}{\left(3i\right)\left(-3i\right)}\\ &= \frac{6i + 15i^2}{-9i^2}\\ &= \frac{6i + 15\left(-1\right)}{-9\left(-1\right)}\\ &= \frac{6i - 15}{9}\\ &= \frac{-15}{9} + \frac{6}{9}i\\ &= \boxed{~-\frac{5}{3} + \frac{2}{3}i~} \end{align}\]

Division Practice

Try It! 6: Simplify \(\displaystyle{\frac{3 + i}{2 - i}}\)

Try It! 7: Simplify \(\displaystyle{\frac{-7i}{3 + 4i}}\)

Try It! 8: Simplify \(\displaystyle{\frac{3 + 4i}{-7i}}\)

Exit Ticket Task

Navigate to our MAT142 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 9. Complex Numbers

Task: Simplify \(\displaystyle{\left(2 + 5i\right)\left(1 - 6i\right)}\). Express your answer in the form \(a + bi\).

Summary and Next Time…

Ideas From Today

A complex number has the form \(a + bi\) where \(a, b \in \mathbb{R}\) and \(i = \sqrt{-1}\).
Real numbers are complex numbers with \(b = 0\), so \(\mathbb{R}\) is a subset of \(\mathbb{C}\).
Add/subtract: combine real parts and imaginary parts separately (as like terms).
Multiply: distribute and use \(i^2 = -1\) to simplify.
The complex conjugate of \(a + bi\) is \(a - bi\); the product \(\left(a + bi\right)\left(a - bi\right)\) is always real: \(a^2 + b^2\).
Divide: multiply numerator and denominator by the conjugate of the denominator, then simplify.

Looking Ahead

Complex numbers arise naturally when solving quadratic equations – the quadratic formula can produce square roots of negative numbers, and complex numbers are exactly what we need to make sense of them.
When we study quadratic (and higher-degree polynomial) functions next, we’ll see that complex numbers complete the picture of what roots can look like.

Next Time:
Quadratic Functions and Equations

Homework:
Complete Homework 7 on MyOpenMath