Lesson Objectives

- Review the basic concepts of complex numbers
- Learn how to work with the imaginary unit i
- Learn how to find complex solutions of equations
- Learn how to perform operations with complex numbers
- Learn how to simplify powers of i
- Learn how to plot complex numbers on the complex plane
- Learn how to find the absolute value of a complex number

## Basic Concepts of Complex Numbers

We previously studied the imaginary unit $i$ along with the set of complex numbers in our course. We learned that the imaginary unit $i$ is defined as: $$i=\sqrt{-1}$$ $$i^{2}=-1$$

a is known as the real part of the complex number.

b is known as the imaginary part of the complex number, although in some texts, the term b$i$ is defined to be the imaginary part.

For two complex numbers to be equal, the real parts must be equal and the imaginary parts must be equal: $$a + bi=c + di$$ if and only if: $$a=c$$ $$\text{and}$$ $$b=d$$ The set of real numbers is a subset of the complex numbers. $$a + bi$$ If we let b = 0: $$a + 0i=a$$ Therefore, a real number such as 7 can be written as: $$7 + 0i=7$$ When b ≠ 0, we have a nonreal complex number:

Therefore, 2 + 5$i$ is a nonreal complex number.

When a = 0 and b ≠ 0, then our nonreal complex number is said to be a pure imaginary number. Therefore, a nonreal complex number such as 4$i$ is a pure imaginary number, and can be written as: $$0 + 4i=4i$$ The standard form for a complex number is: $$a + bi$$ When radicals are involved, we may switch the $i$ and b to make things more clear: $$a + ib$$

Example #1: Simplify each. $$\sqrt{-147}$$ We can use our product rule for radicals: $$\sqrt{-147}=\sqrt{-1}\cdot \sqrt{49}\cdot \sqrt{3}$$ $$\sqrt{-1}=i$$ $$\sqrt{49}=7$$ Let's use these facts to simplify our expression: $$\sqrt{-147}=\sqrt{-1}\cdot \sqrt{49}\cdot \sqrt{3}$$ $$\sqrt{-147}=7i \sqrt{3}$$ Example #2: Simplify each. $$\sqrt{-180}$$ We can use our product rule for radicals: $$\sqrt{-180}=\sqrt{-1}\cdot \sqrt{36}\cdot \sqrt{5}$$ $$\sqrt{-1}=i$$ $$\sqrt{36}=6$$ Let's use these facts to simplify our expression: $$\sqrt{-180}=\sqrt{-1}\cdot \sqrt{36}\cdot \sqrt{5}$$ $$\sqrt{-180}=6i \sqrt{5}$$

Example #3: Solve each equation. $$2x^2 + 10x + 5=8x$$ Let's write this in the form of: $$ax^2 + bx + c=0$$ $$2x^2 + 10x + 5=8x$$ Subtract 8x from both sides: $$2x^2 + 2x + 5=0$$ Plug into the quadratic formula: $$a=2$$ $$b=2$$ $$c=5$$ $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-2 \pm \sqrt{2^2 - 4(2)(5)}}{2(2)}$$ $$x=\frac{-2 \pm \sqrt{4 - 40}}{4}$$ $$x=\frac{-2 \pm \sqrt{-36}}{4}$$ $$x=\frac{-2 \pm 6i}{4}$$ $$x=\frac{-1 \pm 3i}{2}$$ Our solution set, written in standard form: $$\left\{-\frac{1}{2}\pm \frac{3}{2}i \right\}$$

Example #4: Simplify each. $$\sqrt{-3}\cdot 2\sqrt{-8}$$ First, simplify each: $$i\sqrt{3}\cdot 4i\sqrt{2}$$ Now, we can perform our multiplication: $$i \cdot 4i \cdot \sqrt{3 \cdot 2}$$ $$4i^{2}\sqrt{6}$$ Use the rule for the imaginary unit squared: $$i^{2}=-1$$ $$4i^{2}\sqrt{6}$$ $$-4\sqrt{6}$$

Example #5: Simplify each. $$(4 + 2i) + (-7 + 7i)$$ Add the real parts, then add the imaginary parts: $$(4 + (-7)) + (2 + 7)i$$ Simplify: $$-3 + 9i$$ Example #6: Simplify each. $$(7 + 2i) - (-5 - i)$$ Subtract the real parts, then subtract the imaginary parts: $$(7 - (-5)) + (2 - (-1))i$$ Simplify: $$12 + 3i$$ It may be easier to convert subtraction into addition of the opposite: $$(7 + 2i) - (-5 - i)$$ Change the minus to plus and the sign of each term inside of the second set of parentheses: $$(7 + 2i) + (5 + i)$$ $$(7 + 5) + (2 + 1)i$$ $$12 + 3i$$

Example #7: Simplify each. $$(-4 - 3i)(2 + 4i)$$ For this problem, we can just use FOIL: $$\text{F}: -4 \cdot 2=-8$$ $$\text{O}: -4 \cdot 4i=-16i$$ $$\text{I}: -3i \cdot 2=-6i$$ $$\text{L}: -3i \cdot 4i=-12i^{2}$$ Putting these parts together, we obtain: $$-8 - 16i - 6i - 12i^{2}$$ $$-8 - 22i - 12(-1)$$ $$-8 - 22i + 12$$ $$4 - 22i$$

Example #8: Simplify each. $$(4 + 3i)(4 - 3i)$$ Let's use the shortcut first and then we can show this using the FOIL method: $$(a + bi)(a - bi)=a^{2}+ b^{2}$$ $$(4 + 3i)(4 - 3i)=25$$ Using FOIL: $$\text{F}: 4 \cdot 4=16$$ $$\text{O}: 4 \cdot -3i=-12i$$ $$\text{I}: 3i \cdot 4=12i$$ $$\text{L}: 3i \cdot -3i=-9i^{2}$$ $$16 - 12i + 12i - 9i^{2}$$ $$16 - 9(-1)$$ $$16 + 9=25$$

Example #9: Simplify each. $$\frac{-3 - 4i}{5 + 4i}$$ To simplify here, we will multiply the numerator and denominator by the complex conjugate of the denominator: $$\frac{-3 - 4i}{5 + 4i}\cdot \frac{5 - 4i}{5 - 4i}=\frac{-31 - 8i}{41}$$ In standard form, our answer is: $$-\frac{31}{41}- \frac{8}{41}i$$

Example #10 Simplify each. $$i^{151}$$ We want to use our rules of exponents to rewrite this problem.

Since $i^{4}$ = 1, let's divide 151 by 4:

151 ÷ 4 = 37 R3

This tells us that 148 is divisible by 4: $$i^{151}=i^{148}\cdot i^{3}$$ $$i^{151}=(i^{4})^{37}\cdot i^{3}$$ $$i^{151}=(1)^{37}\cdot i^{3}$$ $$i^{151}=1 \cdot i^{3}$$ $$i^{151}=i^{3}=-i$$ Example #11: Simplify each. $$i^{-38}$$ We could solve this in two ways: First, let's think about 40, which is the smallest number that is divisible by 4 that is larger than 38. $$i^{40}=(i^{4})^{10}=1^{10}=1$$ We can multiply any number by 1 and it remains unchanged: $$i^{-38}\cdot i^{40}=i^2=-1$$ Another method, which is slower would involve first setting up a fraction: $$i^{-38}=\frac{1}{i^{38}}$$ 38 ÷ 4 = 9 R2 $$\frac{1}{i^{38}}=\frac{1}{(i^{4})^{9}}\cdot \frac{1}{i^{2}}$$ $$\frac{1}{i^{38}}=1 \cdot \frac{1}{i^{2}}$$ $$\frac{1}{i^{38}}=1 \cdot \frac{1}{-1}$$ $$\frac{1}{i^{38}}=-1$$

Example #12: Plot the given complex number. $$5 - 7i$$ To plot this complex number on the complex plane, we will move from the origin 5 units to the right on the real axis since 5 is the real part of our complex number. Next, we will move 7 units down on the imaginary axis since -7 is the imaginary part of our complex number. Lastly, we can label our point as: 5 - 7$i$.

Example #13: Find the absolute value. $$|2 + 6i|$$ To find the absolute value, let's use our formula: $$|2 + 6i|=\sqrt{2^2 + 6^2}$$ $$|2 + 6i|=\sqrt{40}$$ $$|2 + 6i|=2\sqrt{10}$$

### Complex Numbers

A complex number is of the form: $$a + bi$$ Where a and b are real numbers.a is known as the real part of the complex number.

b is known as the imaginary part of the complex number, although in some texts, the term b$i$ is defined to be the imaginary part.

For two complex numbers to be equal, the real parts must be equal and the imaginary parts must be equal: $$a + bi=c + di$$ if and only if: $$a=c$$ $$\text{and}$$ $$b=d$$ The set of real numbers is a subset of the complex numbers. $$a + bi$$ If we let b = 0: $$a + 0i=a$$ Therefore, a real number such as 7 can be written as: $$7 + 0i=7$$ When b ≠ 0, we have a nonreal complex number:

Therefore, 2 + 5$i$ is a nonreal complex number.

When a = 0 and b ≠ 0, then our nonreal complex number is said to be a pure imaginary number. Therefore, a nonreal complex number such as 4$i$ is a pure imaginary number, and can be written as: $$0 + 4i=4i$$ The standard form for a complex number is: $$a + bi$$ When radicals are involved, we may switch the $i$ and b to make things more clear: $$a + ib$$

### Simplifying Expressions Using the Imaginary Unit

For a positive real number a: $$a > 0$$ $$\sqrt{-a}=i\sqrt{a}$$ We can use this property to simplify expressions involving the square root of a negative number. Let's look at a few examples of simplifying expressions using the imaginary unit $i$.Example #1: Simplify each. $$\sqrt{-147}$$ We can use our product rule for radicals: $$\sqrt{-147}=\sqrt{-1}\cdot \sqrt{49}\cdot \sqrt{3}$$ $$\sqrt{-1}=i$$ $$\sqrt{49}=7$$ Let's use these facts to simplify our expression: $$\sqrt{-147}=\sqrt{-1}\cdot \sqrt{49}\cdot \sqrt{3}$$ $$\sqrt{-147}=7i \sqrt{3}$$ Example #2: Simplify each. $$\sqrt{-180}$$ We can use our product rule for radicals: $$\sqrt{-180}=\sqrt{-1}\cdot \sqrt{36}\cdot \sqrt{5}$$ $$\sqrt{-1}=i$$ $$\sqrt{36}=6$$ Let's use these facts to simplify our expression: $$\sqrt{-180}=\sqrt{-1}\cdot \sqrt{36}\cdot \sqrt{5}$$ $$\sqrt{-180}=6i \sqrt{5}$$

### Solving Quadratic Equations with Complex Solutions

Recall that complex numbers allow us to solve many types of equations where the square root of a negative number is involved. Let's look at an example.Example #3: Solve each equation. $$2x^2 + 10x + 5=8x$$ Let's write this in the form of: $$ax^2 + bx + c=0$$ $$2x^2 + 10x + 5=8x$$ Subtract 8x from both sides: $$2x^2 + 2x + 5=0$$ Plug into the quadratic formula: $$a=2$$ $$b=2$$ $$c=5$$ $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-2 \pm \sqrt{2^2 - 4(2)(5)}}{2(2)}$$ $$x=\frac{-2 \pm \sqrt{4 - 40}}{4}$$ $$x=\frac{-2 \pm \sqrt{-36}}{4}$$ $$x=\frac{-2 \pm 6i}{4}$$ $$x=\frac{-1 \pm 3i}{2}$$ Our solution set, written in standard form: $$\left\{-\frac{1}{2}\pm \frac{3}{2}i \right\}$$

### Operations with Radicals

Let's review how to perform operations with radicals such as addition, subtraction, multiplication, and division. When we have a product or quotient with a negative radicand, we will first simplify before performing any operations: $$\sqrt{-a}=i\sqrt{a}$$ Note: When working with negative radicands, we need to be really careful with the product rule for radicals. $$\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$$ if and only if a and b are not both negative. Let's look at an example.Example #4: Simplify each. $$\sqrt{-3}\cdot 2\sqrt{-8}$$ First, simplify each: $$i\sqrt{3}\cdot 4i\sqrt{2}$$ Now, we can perform our multiplication: $$i \cdot 4i \cdot \sqrt{3 \cdot 2}$$ $$4i^{2}\sqrt{6}$$ Use the rule for the imaginary unit squared: $$i^{2}=-1$$ $$4i^{2}\sqrt{6}$$ $$-4\sqrt{6}$$

### Adding and Subtracting Complex Numbers

For the complex numbers a + b$i$ and c + d$i$: $$(a + bi) + (c + di)=(a + c) + (b + d)i$$ $$(a + bi) - (c + di)=(a - c) + (b - d)i$$ In other words, to add or subtract complex numbers, add or subtract the real parts, then add or subtract the imaginary parts. Let's look at a few examples.Example #5: Simplify each. $$(4 + 2i) + (-7 + 7i)$$ Add the real parts, then add the imaginary parts: $$(4 + (-7)) + (2 + 7)i$$ Simplify: $$-3 + 9i$$ Example #6: Simplify each. $$(7 + 2i) - (-5 - i)$$ Subtract the real parts, then subtract the imaginary parts: $$(7 - (-5)) + (2 - (-1))i$$ Simplify: $$12 + 3i$$ It may be easier to convert subtraction into addition of the opposite: $$(7 + 2i) - (-5 - i)$$ Change the minus to plus and the sign of each term inside of the second set of parentheses: $$(7 + 2i) + (5 + i)$$ $$(7 + 5) + (2 + 1)i$$ $$12 + 3i$$

### Multiplying Complex Numbers

The product of two complex numbers can be found using our distributive property and the definition for $i$ squared. Let's look at an example.Example #7: Simplify each. $$(-4 - 3i)(2 + 4i)$$ For this problem, we can just use FOIL: $$\text{F}: -4 \cdot 2=-8$$ $$\text{O}: -4 \cdot 4i=-16i$$ $$\text{I}: -3i \cdot 2=-6i$$ $$\text{L}: -3i \cdot 4i=-12i^{2}$$ Putting these parts together, we obtain: $$-8 - 16i - 6i - 12i^{2}$$ $$-8 - 22i - 12(-1)$$ $$-8 - 22i + 12$$ $$4 - 22i$$

### Multiplying Complex Conjugates

When we have two binomials with identical terms but opposite signs, we have conjugates. $$a + bi$$ $$a - bi$$ The two complex numbers above are known as complex conjugates. The product of complex conjugates is always a real number. $$(a + bi)(a - bi)=a^{2}+ b^{2}$$ Let's look at an example.Example #8: Simplify each. $$(4 + 3i)(4 - 3i)$$ Let's use the shortcut first and then we can show this using the FOIL method: $$(a + bi)(a - bi)=a^{2}+ b^{2}$$ $$(4 + 3i)(4 - 3i)=25$$ Using FOIL: $$\text{F}: 4 \cdot 4=16$$ $$\text{O}: 4 \cdot -3i=-12i$$ $$\text{I}: 3i \cdot 4=12i$$ $$\text{L}: 3i \cdot -3i=-9i^{2}$$ $$16 - 12i + 12i - 9i^{2}$$ $$16 - 9(-1)$$ $$16 + 9=25$$

### Dividing Complex Numbers

When we end up with a radical in the denominator, we want to rationalize the denominator. This means if $i$ is in any denominator, we will want to rationalize the denominator as it represents the square root of -1. Let's look at an example.Example #9: Simplify each. $$\frac{-3 - 4i}{5 + 4i}$$ To simplify here, we will multiply the numerator and denominator by the complex conjugate of the denominator: $$\frac{-3 - 4i}{5 + 4i}\cdot \frac{5 - 4i}{5 - 4i}=\frac{-31 - 8i}{41}$$ In standard form, our answer is: $$-\frac{31}{41}- \frac{8}{41}i$$

### Simplifying Powers of $i$

We can use a simple pattern to simplify powers of the imaginary unit $i$: $$i^{1}=i$$ $$i^{2}=-1$$ $$i^{3}=i \cdot i^2=-i$$ $$i^{4}=i^{2}\cdot i^{2}=1$$ Let's look at a few examples.Example #10 Simplify each. $$i^{151}$$ We want to use our rules of exponents to rewrite this problem.

Since $i^{4}$ = 1, let's divide 151 by 4:

151 ÷ 4 = 37 R3

This tells us that 148 is divisible by 4: $$i^{151}=i^{148}\cdot i^{3}$$ $$i^{151}=(i^{4})^{37}\cdot i^{3}$$ $$i^{151}=(1)^{37}\cdot i^{3}$$ $$i^{151}=1 \cdot i^{3}$$ $$i^{151}=i^{3}=-i$$ Example #11: Simplify each. $$i^{-38}$$ We could solve this in two ways: First, let's think about 40, which is the smallest number that is divisible by 4 that is larger than 38. $$i^{40}=(i^{4})^{10}=1^{10}=1$$ We can multiply any number by 1 and it remains unchanged: $$i^{-38}\cdot i^{40}=i^2=-1$$ Another method, which is slower would involve first setting up a fraction: $$i^{-38}=\frac{1}{i^{38}}$$ 38 ÷ 4 = 9 R2 $$\frac{1}{i^{38}}=\frac{1}{(i^{4})^{9}}\cdot \frac{1}{i^{2}}$$ $$\frac{1}{i^{38}}=1 \cdot \frac{1}{i^{2}}$$ $$\frac{1}{i^{38}}=1 \cdot \frac{1}{-1}$$ $$\frac{1}{i^{38}}=-1$$

### Complex Plane (Argand Diagram)

Lastly, let's review how to plot complex numbers on the complex plane. A complex plane, also known as an Argand diagram looks the same as our real number coordinate plane. The only difference is that we have renamed our x-axis as the real axis and our y-axis as the imaginary axis. If we want to plot a complex number (a + b$i$) on the complex plane, we move from the origin a units horizontally and b units vertically. Let's look at an example.Example #12: Plot the given complex number. $$5 - 7i$$ To plot this complex number on the complex plane, we will move from the origin 5 units to the right on the real axis since 5 is the real part of our complex number. Next, we will move 7 units down on the imaginary axis since -7 is the imaginary part of our complex number. Lastly, we can label our point as: 5 - 7$i$.

### Absolute Value of a Complex Number

The absolute value of a complex number is a measure of its distance from zero or the origin (0 + 0i). Since we are working with the origin as one point, our distance formula becomes much easier to work with: $$d=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d=\sqrt{(x_2 - 0)^2 + (y_2 - 0)^2}$$ $$d=\sqrt{x^2 + y^2}$$ If we translate this into the language of complex numbers: $$|a + bi|=\sqrt{a^2 + b^2}$$ Let's look at an example.Example #13: Find the absolute value. $$|2 + 6i|$$ To find the absolute value, let's use our formula: $$|2 + 6i|=\sqrt{2^2 + 6^2}$$ $$|2 + 6i|=\sqrt{40}$$ $$|2 + 6i|=2\sqrt{10}$$

#### Skills Check:

Example #1

Simplify each. $$(-3 - 3i) + (1 - 4i)$$

Please choose the best answer.

A

$$3 + i$$

B

$$-2 + i$$

C

$$4 - i$$

D

$$-2 - 7i$$

E

$$-7i$$

Example #2

Simplify each. $$(2 - i)(-3 + 3i)$$

Please choose the best answer.

A

$$3 - 9i$$

B

$$9 - 3i$$

C

$$-3 + 9i$$

D

$$5 + 4i$$

E

$$5 - 4i$$

Example #3

Simplify each. $$\frac{-1 - 4i}{-1 + 3i}$$

Please choose the best answer.

A

$$-\frac{11}{10}+ \frac{7}{10}i$$

B

$$-\frac{1}{5}- \frac{4}{5}i$$

C

$$-\frac{14}{13}- \frac{5}{13}i$$

D

$$-\frac{1}{10}+ \frac{7}{10}i$$

E

$$\frac{7}{10}- \frac{2}{5}i$$

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