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2. Quadratic Functions
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Algebra 2

2.04 Complex numbers and operations

Lesson
Practice

The imaginary unit

We have found that quadratic equations can have non-real solutions, when the discriminant is less than zero. To define these non-real solutions, we use imaginary numbers which are built on the concept that there is a number, the imaginary unit \left(i\right) such that i^2=-1.

Imaginary number

A number that, when squared, results in a negative real number. It can be expressed as a real multiple of i, defined by i = \sqrt{-1}.

This is not a real number, since the square of any real number is always non-negative. However, using this new number, we can find the answers to many real-world problems in electrical engineering, quantum physics, and more.

Note that the words “real” and “imaginary” are just names for different kinds of numbers. It does not mean that one type of number exists while the other type does not exist.

We can express the square root of any negative number by taking out a factor of \sqrt{-1}=i to get a imaginary number. For example:\sqrt{-5}=\sqrt{-1\cdot 5}=\sqrt{-1}\cdot\sqrt{5}=i\sqrt{5}

Exploration

Calculate the powers of i, from i^2 until i^8.

  1. What pattern do you notice?
  2. How can you use the pattern to find i^{23},\,i^{30},\,i^{33}, and i^{40}?

When we raise i to some power, we notice a pattern unfolds:

i^1=ii^5=ii^9=i
i^2=-1i^6=-1i^{10}=-1
i^3=-ii^7=-ii^{11}=-i
i^4=1i^8=1i^{12}=1

We can see that there is a cycle of 4, and we can use this pattern to determine the value of any power of i.

  • When the exponent of i is a multiple of 4, the expression simplifies to 1.

  • When the exponent is one more than a multiple of 4, the expression simplifies to i.

  • When the exponent is 2 more than a multiple of 4, the expression simplifies to -1.

  • When the exponents is 3 more then a multiple of 4, the expression simplifies to -i.

We can summarize this algebraically like so:

i^{4n+1}=i
i^{4n+2}=-1
i^{4n+3}=-i
i^{4n}=1

This shows that we can write any exponent in the form {4n+r} where n is any integer and r can be 0,\,1,\, 2,\, or 3.

Examples

Example 1

Express the following in terms of i:

a

\sqrt{-49}

Worked Solution
Create a strategy

If we use the multiplication property of radicals, we can separate the radicand into -1\cdot 49 to define with i.

Apply the idea
\displaystyle \sqrt{-49}\displaystyle =\displaystyle \sqrt{-1\cdot 49}Factor out -1
\displaystyle =\displaystyle \sqrt{-1}\cdot\sqrt{49}Product of square roots
\displaystyle =\displaystyle i\cdot 7Definition of i
\displaystyle =\displaystyle 7iCommutative property of multiplication
b

\sqrt{-10}

Worked Solution
Create a strategy

We will use the same reasoning as the example above.

Apply the idea
\displaystyle \sqrt{-10}\displaystyle =\displaystyle \sqrt{-1\cdot 10}Factor out -1
\displaystyle =\displaystyle \sqrt{-1}\cdot\sqrt{10}Product of square roots
\displaystyle =\displaystyle i\sqrt{10}Definition of i

Since the radical cannot be simplified further, this is the final answer.

Reflect and check

This answer can also be written as \sqrt{10}i, but it needs to be clear that the i is not underneath the radical.

c

\sqrt{-32}

Worked Solution
Create a strategy

After writing the expression in terms of i, this radical can be simplified which we can do using properties of radicals. Our goal is to find the largest perfect square factor of 32.

Apply the idea
\displaystyle \sqrt{-32}\displaystyle =\displaystyle \sqrt{-1\cdot 32}Factor out -1
\displaystyle =\displaystyle \sqrt{-1}\cdot \sqrt{32}Product of radicals
\displaystyle =\displaystyle i\sqrt{32}Substitute i=\sqrt{-1}
\displaystyle =\displaystyle i\sqrt{16\cdot 2}Factor 32
\displaystyle =\displaystyle i\sqrt{16}\sqrt{2}Product of radicals
\displaystyle =\displaystyle 4i\sqrt{2}Evaluate the radical
Reflect and check

As you become more familiar with imaginary numbers, you can use fewer steps to simplify:

\displaystyle \sqrt{-32}\displaystyle =\displaystyle \sqrt{-16\cdot 2}
\displaystyle =\displaystyle \sqrt{-16}\cdot\sqrt{2}
\displaystyle =\displaystyle 4i\sqrt{2}

Example 2

Simplify the following expressions:

a

i^{15}

Worked Solution
Create a strategy

We can use properties of exponents and the fact that i^4=1 to make the process simpler. If we can find out how many times 4 goes into 15, then we only need to know the first 3 powers of i to simplify.

Apply the idea

Using the fact that 15=4\cdot 3+3,

\displaystyle i^{15}\displaystyle =\displaystyle i^{4\cdot 3+3}Rewrite the exponent
\displaystyle =\displaystyle (i^4)^3\cdot i^3Rewrite using power and product property
\displaystyle =\displaystyle (1)^3\cdot -iSubstitute i^4=1 and i^3=-i
\displaystyle =\displaystyle -iEvaluate the exponent

Therefore, i^{15}=-i.

Reflect and check

We can use fewer steps by dividing the exponent by 4 and determining the remainder.

15\div4=3 with a remainder of 3

The remainder will be the new exponent of i that we will use to evaluate the expression.

i^3=-i which means i^{15}=-i.

b

i^{-6}

Worked Solution
Create a strategy

The definition of negative exponents says i^{-m}=\dfrac{1}{i^m}.

Apply the idea
\displaystyle i^{-6}\displaystyle =\displaystyle \dfrac{1}{i^6}Definition of negative exponents
\displaystyle =\displaystyle \dfrac{1}{i^4\cdot i^2}Product of powers
\displaystyle =\displaystyle \dfrac{1}{1\cdot -1}Substitute i^4=1 and i^2=-1
\displaystyle =\displaystyle -1Evaluate the multiplication and division

Therefore, i^{-6}=-1

Reflect and check

If we extend the cyclic pattern into the negative exponents, we see that this still follows the pattern i,\,-1,\,-i,\,1:

i^{-7}i^{-6}i^{-5}i^{-4}i^{-3}i^{-2}i^{-1}i^{0}i^1i^2i^3i^4
i-1-i1i-1-i1i-1-i1
c

5 i^{2} - 2 i^{4} + 3 i^{7}

Worked Solution
Create a strategy

We have already calculated these powers of i above:

i^2=-1

i^4=1

i^7=-i

Apply the idea
\displaystyle 5 i^{2} - 2 i^{4} + 3 i^{7}\displaystyle =\displaystyle 5\left(-1\right)-2\left(1\right)+3\left(-i\right)Substitute i^2=-1, i^4=1, and i^7=-1
\displaystyle =\displaystyle -5-2-3iEvaluate the multiplication
\displaystyle =\displaystyle -7-3iEvaluate the subtraction
Reflect and check

Because one term has an i and one does not, these are not like terms and cannot be combined.

Idea summary

Imaginary numbers are used to define square roots of negative numbers. The imaginary unit i is defined as \sqrt{-1}. The powers of i follow a cyclic pattern of 4:

i^1i^2i^3i^4\cdots
i-1-i1\cdots

Complex numbers

Previously, all the numbers we knew fell under the umbrella of real numbers. Now, we have learned about imaginary numbers which are not real numbers, so we need to introduce a new type of number that encompasses both real and imaginary numbers.

Complex number

The set of all real and imaginary numbers.

a \pm bi where a and b are real numbers, and i=\sqrt{-1}.

A complex number consists of both real (a) and imaginary (bi) but either part can be 0.

By this definition, all numbers are complex numbers. Sometimes, a is referred to as the real part and bi is called the imaginary part. Both parts together make up one complex number.

A real number is a complex number where b=0. For example, -5=-5+0i.

A pure imaginary number is a complex number where a=0. For example, 2i=0+2i.

A diagram showing the Complex Number System. Ask your teacher for more information.

We perform the algebraic operations for complex numbers the same way we perform operations for rational algebraic expressions, except we sometimes have an extra step to account for the powers of i.

Addition: add in the same way as binomials with like terms

General example\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i
Numerical example\left(2+3i\right)+\left(4+5i\right)=\left(2+4\right)+\left(3+5\right)i=6+8i

Subtraction: subtract in the same way as binomials with like terms

General example\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i
Numerical example\left(2+3i\right)-\left(4+5i\right)=\left(2-4\right)+\left(3-5\right)i=-2-2i

Multiplication: distribute in the same way as binomials, evaluate any powers of i, then combine any like terms

General example\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bdi^2=ac+\left(ad+bc\right)i+bdi^2=\left(ac-bd\right)+\left(ad+bc\right)i
Numerical example\left(2+3i\right)\left(4+5i\right)=8+10i+12i+15i^2=8+\left(10+12\right)i+15i^2={\left(8-15\right)+\left(10+12\right)i}=-7+22i

The conjugate of the complex number 𝑎+𝑏𝑖 is 𝑎−𝑏𝑖.

A complex number multiplied by its conjugate is a non-negative, real number.

General example\left(a+bi\right)\left(a-bi\right)=a^2-abi+abi-b^2i^2=a^2+b^2i^2=a^2-b^2
Numerical example\left(5+2i\right)\left(5-2i\right)=5^2-10i+10i-4i^2=25-4i^2=25+4=29

Examples

Example 3

The following complex numbers are written in the form a+bi. State the values a and b.

a

-8+i

Worked Solution
Apply the idea

a=-8

b=1

Reflect and check

Both terms together are considered one complex number.

b

\dfrac{4}{7}

Worked Solution
Create a strategy

This is a real number which means the imaginary part is missing, so we can write the imaginary part as 0i. That means \dfrac{4}{7} can be written as \dfrac{4}{7}+0i.

Apply the idea

a=\dfrac{4}{7} and b=0

Reflect and check

\dfrac{4}{7} can be classified as complex, real, and rational.

c

\sqrt{-40}

Worked Solution
Create a strategy

We need to rewrite the expression in terms of i, simplify the radical, then determine the values of a and b.

Apply the idea
\displaystyle \sqrt{-40}\displaystyle =\displaystyle \sqrt{-1\cdot 40}Factor out -1
\displaystyle =\displaystyle \sqrt{-1}\sqrt{40}Product of radicals
\displaystyle =\displaystyle i\sqrt{40}Definition of i
\displaystyle =\displaystyle i\sqrt{4\cdot 10}Factor 40
\displaystyle =\displaystyle i\sqrt{4}\sqrt{10}Product of radicals
\displaystyle =\displaystyle 2i\sqrt{10}Evaluate the square root

There is only one term in this expression, and it contains i. This means the real part is missing which makes this a pure imaginary number.

2i\sqrt{10}=0+2\sqrt{10}i

a=0

b=2\sqrt{10}

Reflect and check

This number can be classified as complex, imaginary, and pure imaginary.

Example 4

Simplify each of the following expressions. Justify each step using the commutative, associative, and distributive properties.

a

\left( -3 + 5 i\right) + \left(7-2 i\right)

Worked Solution
Create a strategy

When adding or subtracting complex numbers, we can combine like terms by adding the real parts together and the imaginary parts together.

Apply the idea
\displaystyle \left( -3 + 5 i\right) + \left(7-2 i\right)\displaystyle =\displaystyle (-3 +7)+ (5 i -2 i)Commutative and associative properties
\displaystyle =\displaystyle 4+3iEvaluate the addition
b

(-6-i)-(8-5i)

Worked Solution
Apply the idea
\displaystyle (-6-i)-(8-5i)\displaystyle =\displaystyle (-6-i)+(-8+5i)Distributive property
\displaystyle =\displaystyle (-6+-8)+(-i+5i)Commutative and associative properties
\displaystyle =\displaystyle -14+4iEvaluate the addition
Reflect and check

It is important to distribute the negative sign to the second complex number first because the commutative and associative properties only hold for addition and multiplication, not subtraction.

c

\left( 2 - 4 i\right) \left(-4+2 i\right)

Worked Solution
Create a strategy

When multiplying complex numbers, we can use the distributive property: \left(A+B\right)\left(C+D\right)=AC+AD+BC+BD

Apply the idea
\displaystyle \left( 2 - 4 i\right) \left(-4+2 i\right)\displaystyle =\displaystyle 2\left(-4\right)+2\left(2i\right)-4i\left(-4\right)-4i\left(2i\right)Distributive property
\displaystyle =\displaystyle -8+4i+16i-8i^2Evaluate the multiplication
\displaystyle =\displaystyle -8+4i+16i-8(-1)Substitute i^2=-1
\displaystyle =\displaystyle -8+4i+16i+8Evaluate the multiplication
\displaystyle =\displaystyle 20iEvaluate the addition

Example 5

What is the result when each of the following is multiplied by its conjugate?

a

4+3i

Worked Solution
Create a strategy

First, we need to identify the conjugate of the expression. The conjugate of 4+3i is 4-3i.

When multiplying complex numbers, we can use the distributive property: \left(A+B\right)\left(C+D\right)=AC+AD+BC+BD

Apply the idea
\displaystyle \left( 4 + 3 i\right) \left(4-3 i\right)\displaystyle =\displaystyle 4\left(4\right)+4\left(-3i\right)+3i\left(4\right)+3i\left(-3i\right)Distributive property
\displaystyle =\displaystyle 16-12i+12i-9i^2Evaluate the multiplication
\displaystyle =\displaystyle 16-12i+12i-9(-1)Substitute i^2=-1
\displaystyle =\displaystyle 16-12i+12i+9Evaluate the multiplication
\displaystyle =\displaystyle 25Evaluate the addition
Reflect and check

Recall that the product of complex conjugates always results in a real number. We can use this fact to check the reasonableness our solution. If the solution is imaginary, then there must be a mistake in the calculations.

b

-\sqrt{-4}

Worked Solution
Create a strategy

First, we need to rewrite the expression as a complex number. Then, we need to identify the conjugate of the expression.

Apply the idea

As a complex number, the expression is:

\displaystyle -\sqrt{-4}\displaystyle =\displaystyle -\sqrt{-1\cdot 4}Rewrite as a product with -1
\displaystyle =\displaystyle -\sqrt{-1}\sqrt{4}Product of radicals
\displaystyle =\displaystyle -i\sqrt{4}Definition of i
\displaystyle =\displaystyle -2iEvaluate the square root

This is a pure imaginary number because the real part is 0, so we can also write this as 0-2i. This helps us see that the conjugate of -2i is 2i.

\displaystyle \left(-2i\right) \left(2i\right)\displaystyle =\displaystyle -2\left(2\right)i\left(i\right)Distributive property
\displaystyle =\displaystyle -4i^2Evaluate the multiplication
\displaystyle =\displaystyle -4(-1)Substitute i^2=-1
\displaystyle =\displaystyle 4Evaluate the multiplication
c

\dfrac{5i-6}{12}

Worked Solution
Create a strategy

First, let's write the expression in the form a+bi.

\displaystyle \dfrac{5i-6}{12}\displaystyle =\displaystyle \dfrac{5i}{12}-\dfrac{6}{12}
\displaystyle =\displaystyle -\dfrac{6}{12}+\dfrac{5i}{12}

Next, we need to identify the conjugate of the expression.

\displaystyle -\dfrac{6}{12}-\dfrac{5i}{12}\displaystyle =\displaystyle \dfrac{-6-5i}{12}
\displaystyle =\displaystyle \dfrac{-5i-6}{12}

The conjugate of \dfrac{5i-6}{12} is \dfrac{-5i-6}{12}.

Apply the idea
\displaystyle \frac{(5i-6)}{12}\cdot\dfrac{(-5i-6)}{12}\displaystyle =\displaystyle \frac{(5i-6)(-5i-6)}{144}Multiply the fractions
\displaystyle =\displaystyle \frac{5i(-5i)+5i(-6)-6(-5i)-6(-6)}{144}Distributive property
\displaystyle =\displaystyle \frac{-25i^2-30i+30i+36}{144}Evaluate the multiplication
\displaystyle =\displaystyle \frac{-25(-1)-30i+30i+36}{144}Substitute i^2=-1
\displaystyle =\displaystyle \frac{25-30i+30i+36}{144}Evaluate the multiplication
\displaystyle =\displaystyle \frac{61}{144}Evaluate the addition
Reflect and check

The only difference between a complex number and its conjugate is the sign of the imaginary parts. The signs of the real parts of complex conjugates are the same, but the signs of the imaginary parts are opposite.

This helps us see that the conjugate of \dfrac{5i-6}{12} is not \dfrac{5i+6}{12} because the signs of the real parts of the expression are different.

Idea summary

All numbers are complex numbers. They are in the form a+bi where a is the real part and bi is the imaginary part. Both parts together make one complex number.

We simplify complex numbers by combining the real parts with the imaginary parts. Sometimes, we have to evaluate powers of i and continue simplifying the expression.

The conjugate of the complex number 𝑎+𝑏𝑖 is 𝑎−𝑏𝑖.

Outcomes

A2.EO.4

The student will perform operations on complex numbers.

A2.EO.4a

Explain the meaning of i.

A2.EO.4b

Identify equivalent radical expressions containing negative rational numbers and expressions in a + bi form.

A2.EO.4c

Apply properties to add, subtract, and multiply complex numbers.

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