Lesson Objectives

- Learn about the imaginary unit i
- Learn how to find the square root of a negative number
- Learn about complex numbers
- Learn how to perform operations with complex numbers
- Learn how to rationalize imaginary denominators

## The Imaginary Unit i and Complex Numbers

Up to this point, we have not been able to find the solution for a problem in which we had to take the square root of a negative number. The problem occurs
because squaring a number always yields a non-negative result. Let's suppose we saw the following equation:
$$\require{color}x^2 = -9$$
We know that there is no real number that can be squared and give a result of -9. To deal with this situation, we will introduce the imaginary unit "i".
The imaginary unit "i" is a number whose square is (-1).
$$i^2 = -1$$
If we take the square root of each side:
$$\sqrt{i^2} = \sqrt{-1}$$
$$i = \sqrt{-1}$$
We can define i as the square root of -1. We can use this property to deal with our equation above.
$$x^2 = -9$$
Take the square root of each side:
$$\sqrt{x^2} = \sqrt{-9}$$
$$x = \sqrt{-9}$$
Let's use i to solve our problem.
$$x = \sqrt{-1} \cdot \sqrt{9}$$
We know the square root of (-1) is defined as i so we can replace this in our problem:
$$x = i\sqrt{9}$$
Since the square root of 9 is 3, we can write our answer as:
$$x = 3i$$
Does this solution actually work in the original equation? Let's try it out.
$$x^2 = -9$$
$$(3i)^2 = -9$$
$$3^2 \cdot i^2 = -9$$
Remember i squared is defined as (-1).
$$9(-1) = -9$$
$$-9 = -9 \hspace{.2em} \color{green}{✔}$$
It's worth noting that -3i works as a solution as well. When solving this equation, we would take the principal and negative square root of
(-9). We will learn this rule in our next lesson when we talk about the square root property.
Let's look at a few examples.

Example 1: Write each as the product of i and a real number $$\sqrt{-5}$$ Let's begin by writing the square root of -5 as the square root of -1 times the square root of 5. $$\sqrt{-1} \cdot \sqrt{5}$$ We can replace the square root of (-1) with i. $$i\sqrt{5}$$ Example 2: Write each as the product of i and a real number $$\sqrt{-44}$$ Let's begin by writing the square root of -44 as the square root of -1 times the square root of 4 times the square root of 11 $$\sqrt{-1} \cdot \sqrt{4} \cdot \sqrt{11}$$ We can replace the square root of (-1) with i and the square root of 4 with 2. $$2i\sqrt{11}$$

Example 3: Simplify each $$\sqrt{-2} \cdot \sqrt{-14}$$ First, we will change our form. $$i\sqrt{2} \cdot i\sqrt{14}$$ Now we can use our product rule for radicals. $$i\sqrt{2} \cdot i\sqrt{14} = i^2\sqrt{4 \cdot 7} = -2\sqrt{7}$$ This method will also apply to a quotient. Let's look at an example.

Example 4: Simplify each $$\frac{\sqrt{-75}}{\sqrt{-3}}$$ First, we will change our form. $$\frac{i\sqrt{75}}{i\sqrt{3}}$$ Let's cancel common factors. $$\require{cancel}\frac{i \cdot \sqrt{3} \cdot \sqrt{25}}{i \cdot \sqrt{3}} = \frac{\cancel{i} \cdot \cancel{\sqrt{3}} \cdot \sqrt{25}}{\cancel{i} \cdot \cancel{\sqrt{3}}} = 5$$

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 5: Simplify each $$(1 + 2i) + (-6 + 9i) - (-9 + 2i)$$ 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. $$(1 - 6 + 9) + (2i + 9i - 2i) = 4 + (2 + 9 - 2)i = 4 + 9i$$ We may also run into multiplication and division with complex numbers. Let's look at an example of multiplication.

Example 6: 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 = -24i

If we simplify:

-21 - 9i - 56i - 24i

Note: Remember i

Example 7: 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 i

(a + bi)(a - bi) = a

Why is there a plus instead of a minus? This is from the i

Example 8: 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. $$\frac{17 - 11i}{10} = \frac{17}{10} - \frac{11i}{10}$$

Example 9: Simplify each $$i^{37}$$ Since 36 is divisible by 4, we will use the product rule for exponents to rewrite our problem as: $$i^{37} = i^{36} \cdot i$$ Now we can use our power to power rule to rewrite the problem as: $$(i^4)^9 \cdot i$$ We know that i

Example 1: Write each as the product of i and a real number $$\sqrt{-5}$$ Let's begin by writing the square root of -5 as the square root of -1 times the square root of 5. $$\sqrt{-1} \cdot \sqrt{5}$$ We can replace the square root of (-1) with i. $$i\sqrt{5}$$ Example 2: Write each as the product of i and a real number $$\sqrt{-44}$$ Let's begin by writing the square root of -44 as the square root of -1 times the square root of 4 times the square root of 11 $$\sqrt{-1} \cdot \sqrt{4} \cdot \sqrt{11}$$ We can replace the square root of (-1) with i and the square root of 4 with 2. $$2i\sqrt{11}$$

### Multiplying Square Roots of Negative Numbers

Up to this point, we have been using the product rule for radicals to simplify. When square roots of negative numbers are involved, we can't use the same rule. We have to change the form first before we can multiply. To see this in action, let's suppose we had the following problem: $$\sqrt{-6} \cdot \sqrt{-3}$$ Based on our prior experience, one may try to solve the problem in this manner: $$\sqrt{-6} \cdot \sqrt{-3} = \sqrt{-6 \cdot -3} = \sqrt{18} = 3\sqrt{2} \hspace{.2em} \color{red}{✖}$$ The product rule for radicals is only valid for real numbers. To solve this problem, we can change the form first: $$\sqrt{-6} \cdot \sqrt{-3} = i\sqrt{6} \cdot i\sqrt{3} = i^2\sqrt{18} = -1 \cdot 3\sqrt{2} = -3\sqrt{2}$$ Let's look at an example.Example 3: Simplify each $$\sqrt{-2} \cdot \sqrt{-14}$$ First, we will change our form. $$i\sqrt{2} \cdot i\sqrt{14}$$ Now we can use our product rule for radicals. $$i\sqrt{2} \cdot i\sqrt{14} = i^2\sqrt{4 \cdot 7} = -2\sqrt{7}$$ This method will also apply to a quotient. Let's look at an example.

Example 4: Simplify each $$\frac{\sqrt{-75}}{\sqrt{-3}}$$ First, we will change our form. $$\frac{i\sqrt{75}}{i\sqrt{3}}$$ Let's cancel common factors. $$\require{cancel}\frac{i \cdot \sqrt{3} \cdot \sqrt{25}}{i \cdot \sqrt{3}} = \frac{\cancel{i} \cdot \cancel{\sqrt{3}} \cdot \sqrt{25}}{\cancel{i} \cdot \cancel{\sqrt{3}}} = 5$$

### 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 5: Simplify each $$(1 + 2i) + (-6 + 9i) - (-9 + 2i)$$ 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. $$(1 - 6 + 9) + (2i + 9i - 2i) = 4 + (2 + 9 - 2)i = 4 + 9i$$ We may also run into multiplication and division with complex numbers. Let's look at an example of multiplication.

Example 6: 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 = -24i

^{2}If we simplify:

-21 - 9i - 56i - 24i

^{2}= -21 - 65i + 24 = 3 - 65iNote: Remember i

^{2}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 7: 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 i

^{2}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) = a

^{2}+ b^{2}Why is there a plus instead of a minus? This is from the i

^{2}that occurs, it changes the sign from a minus into a plus. Let's look at an example.Example 8: 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. $$\frac{17 - 11i}{10} = \frac{17}{10} - \frac{11i}{10}$$

### Simplifying Powers of i

Additionally, we may be asked to simplify powers of i. We will use the fact that i^{2}is (-1) along with the rules for exponents to simplify powers of i. We should note the first few powers of i: $$i^1 = i$$ $$i^2 = -1$$ $$i^3 = i^2 \cdot i = -1 \cdot i = -i$$ $$i^4 = i^2 \cdot i^2 = -1 \cdot -1 = 1$$ Let's take a look at an example.Example 9: Simplify each $$i^{37}$$ Since 36 is divisible by 4, we will use the product rule for exponents to rewrite our problem as: $$i^{37} = i^{36} \cdot i$$ Now we can use our power to power rule to rewrite the problem as: $$(i^4)^9 \cdot i$$ We know that i

^{4}is 1, we can replace this in our problem: $$(i^4)^9 \cdot i = 1^9 \cdot i = 1 \cdot i = i$$
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