Lesson Objectives
• Learn how to perform operations with complex numbers
• Learn how to rationalize imaginary denominators

## How to Perform Operations with Complex Numbers

### How to Add and Subtract Complex Numbers

Up to this point, all numbers that we have worked with were real numbers. A complex number is formed using the imaginary unit i along with our real numbers.
If we let "a" and "b" be any two real numbers then:
a + bi is a complex number.
The "a" is the real part, whereas the "bi" is the complex part. We can perform operations with complex numbers using the same tools we used with real numbers. Let's look at an example that involves addition and subtraction.
Example 1: Simplify each $$\require{cancel}(1 + 2i) + (-6 + 9i) - (-9 + 2i)$$ We will first change the subtraction operation into addition of the opposite. $$(1 + 2i) + (-6 + 9i) + (9 - 2i)$$ Now we can use our commutative, associative, and distributive properties to find the sum. We want to add the real parts and imaginary parts separately. $$(1 - 6 + 9) + (2i + 9i - 2i)=4 + (2 + 9 - 2)i=4 + 9i$$

### How to Multiply and Divide Complex Numbers

We may also run into multiplication and division with complex numbers. Let's look at an example of multiplication.
Example 2: Simplify each $$(3 + 8i)(-7 - 3i)$$ For this problem, we can use FOIL
F » 3 • -7 = -21
O » 3 • -3i = -9i
I » 8i • -7 = -56i
L » 8i • -3i = -24i2
If we simplify:
-21 - 9i - 56i - 24i2 = -21 - 65i + 24 = 3 - 65i
Note: Remember i2 is (-1). $$(3 + 8i)(-7 - 3i)=3 - 65i$$

### Rationalizing Imaginary Denominators

When we start dividing with complex numbers, we will run into problems where we have i in the denominator. Since i by definition is the square root of (-1), it represents a radical and can't be in the denominator of a simplified radical expression. Let's look at an example.
Example 3: Simplify each $$\frac{1}{4i}$$ Remember i represents the square root of (-1). $$\frac{1}{4 \cdot \sqrt{-1}}$$ We can multiply both numerator and denominator by the square root of (-1) or i. We know that i2 is (-1): $$\frac{1}{4i}\cdot \frac{i}{i}=\frac{1}{-1 \cdot 4}=-\frac{i}{4}$$ Another scenario that will occur is trying to rationalize a complex denominator with two terms. Recall when we faced this problem with real numbers, we multiplied the numerator and denominator of the fraction by the conjugate of the denominator. We will use the same technique here. When we multiply complex conjugates together, the result is a real number.
(a + bi)(a - bi) = a2 + b2
Why is there a plus instead of a minus? This is from the i2 that occurs, it changes the sign from a minus into a plus. Let's look at an example.
Example 4: Simplify each $$\frac{10 + 8i}{2 + 6i}$$ We can simplify by multiplying both the numerator and the denominator by the complex conjugate of the denominator. $$\frac{10 + 8i}{2 + 6i}\cdot \frac{2 - 6i}{2 - 6i}=\frac{68 - 44i}{40}$$ We can factor out a 4 from the numerator and denominator and cancel. $$\frac{\cancel{4}(17 - 11i)}{\cancel{4}\cdot 10}=\frac{17 - 11i}{10}$$ We can write this in standard form (a + bi) by splitting up the fraction. $$\frac{17 - 11i}{10}=\frac{17}{10}- \frac{11i}{10}$$

#### Skills Check:

Example #1

Simplify each. $$(-4 - 6i) + (-8 + 3i)$$

A
$$12 + 3i$$
B
$$-12 - 3i$$
C
$$5 + 7i$$
D
$$2i$$
E
$$1 - 4i$$

Example #2

Simplify each. $$(-2 - 5i)(-3 - 4i)$$

A
$$2 + 9i$$
B
$$-1 + 4i$$
C
$$-26 - 7i$$
D
$$-14 + 23i$$
E
$$33 + 44i$$

Example #3

Simplify each. $$\frac{3 + i}{2 + 7i}$$

A
$$\frac{20}{53}+ \frac{13}{53}i$$
B
$$\frac{2i}{53}$$
C
$$5i - 53$$
D
$$\frac{13}{53}- \frac{19}{53}i$$
E
$$9i$$