5.9 Imaginary and Complex Numbers

Learning Outcomes

Rewrite square roots with negative radicands in terms of $i.$
State the real and imaginary parts of a complex number.
Multiply radical expressions with negative radicands.
Add and/or subtract complex numbers, giving the result in the form $a+bi.$
Multiply complex numbers, giving the result in the form $a+bi.$
Simplify whole number powers of $i.$
Divide complex numbers including using complex conjugates, giving the result in the form $a+bi.$

We have noted many times in this chapter that square roots of negative numbers do not exist as real numbers. However, these “non-real numbers” do still have applications in many areas of math and also in the real world in fields such as electrical engineering.

You really need only one new number to start working with the square roots of negative numbers. That number is the square root of $−1,\sqrt{-1}$ . The real numbers are those that can be shown on a number line. When something is not real, we often say it is imaginary. So let us call this new number $i$ and use it to represent the square root of $−1$ .

$i=\sqrt{-1}$

Because $\sqrt{x}\,\cdot \,\sqrt{x}=x$ , we can also see that $\sqrt{-1}\,\cdot \,\sqrt{-1}=-1,$ or $i^2=-1$ . Another way to say this is that $i$ is a solution to the equation $x^2=-1$ which previously did not have solutions.

The number $i$ allows us to work with roots of all negative numbers, not just $\sqrt{-1}$ . We will be using our previous Product Rule for Radicals, $\sqrt{ab}=\sqrt{a}\sqrt{b}$ . Be very careful using this rule with imaginary numbers though! Remember when we introduced the rule, we required that both $\sqrt{a}$ and $\sqrt{b}$ exist, which is not the case in this section! For this section we redefine the product rule to work for imaginary numbers.

product rule for imaginary numbers

For any positive number $a$ , define

$\sqrt{-a}=\sqrt{-1\cdot a}=\sqrt{-1}\cdot\sqrt{a}=i\cdot\sqrt{a}$

Let us try an example of using this rule. If a radicand of a square root is negative, always apply this rule first before any other steps.

Example

Simplify. $\sqrt{-4}$

Show Solution

Use the new product rule to rewrite the radical.

$\sqrt{-4}=\sqrt{-1\cdot 4}=\sqrt{-1}\sqrt{4}=\sqrt{-1}\cdot 2$

Use the definition of $i$ to rewrite $\sqrt{-1}$ as $i$ .

$\sqrt{-1}\cdot 2=2i$

Example

Simplify. $\sqrt{-18}$

Show Solution

Use the new product rule to rewrite the radical.

$\sqrt{-18}=\sqrt{-1\cdot 18}=\sqrt{-1}\sqrt{18}$

Since $18$ is not a perfect square, simplify the radical as done previously in this chapter.

$\sqrt{-1}\sqrt{18}=\sqrt{-1}\sqrt{9}\sqrt{2}=\sqrt{-1}3\sqrt{2}=3\sqrt{2} \; i$

In the following video, we show more examples of how to use imaginary numbers to simplify a square root with a negative radicand.

Rewriting the Square Root of a Negative Number

Rewrite the radical using the rule $\sqrt{-a}=\sqrt{-1}\cdot \sqrt{a}$ .
Rewrite $\sqrt{-1}$ as $i$ and simplify the other radical if possible.

Complex Numbers

COMPLEX NUMBERS

A complex number is any number that can be expressed in the standard form $a+ bi$ where $a$ is the real part and $b$ is the imaginary part. We can say that a complex number is the sum of a real number $a$ and a pure imaginary number $bi.$

For example, $5+2i$ is a complex number. So, too, is $-3+4i\sqrt{3}$ . Here are some examples of complex numbers, with real part and imaginary part labeled. Note that $i$ is not included in the imaginary part.

Complex Number	Real Part	Imaginary Part
$3+7i$	$3$	$7$
$18–32i$	$18$	$−32$
$-\frac{3}{5}+i\sqrt{2}$	$-\frac{3}{5}$	$\sqrt{2}$
$\frac{\sqrt{2}}{2}-\frac{1}{2}i$	$\frac{\sqrt{2}}{2}$	$-\frac{1}{2}$

In a number with a radical as part of $b$ , such as $-\frac{3}{5}+i\sqrt{2}$ above, it is acceptable to write the imaginary unit $i$ in front of the radical. Though writing this number as $-\frac{3}{5}+\sqrt{2} \; i$ is technically correct, it makes it more difficult to tell whether $i$ is inside or outside of the radical. Putting it before the radical, as in $-\frac{3}{5}+i\sqrt{2}$ , clears up any confusion. While usually we will require answers to be in the form of a complex number $a+bi$ , we make an exception if $b$ is a radical. Look at these last two examples.

Number	Complex Form: $a+bi$	Real Part	Imaginary Part
$17$	$17+0i$	$17$	$0$
$−3i$	$0–3i$	$0$	$−3$

By making $b=0$ , any real number can be expressed as a complex number. The real number $a$ is written as $a+0i$ in complex form. Similarly, any pure imaginary number can be expressed as a complex number. By making $a=0$ , any pure imaginary number $bi$ can be written as $0+bi$ in complex form.

Example

Write $83.6$ in the form of a complex number.

Show Solution

Remember that a complex number has the form $a+bi$ . You need to figure out what $a$ and $b$ need to be. Since $83.6$ is a real number, it does not contain any imaginary parts, so the value of $b$ is $0$ .

The answer is $83.6+0i$ .

Example

Write $−3i$ in the form of a complex number.

Show Solution

Remember that a complex number has the form $a+bi$ . You need to figure out what $a$ and $b$ need to be. Since $−3i$ is a pure imaginary number, the value of $a$ is $0$ .

The answer is $0-3i$ .

In practice if the real or imaginary part of a complex number is $0$ we will not write that part at all (so we would leave it as $-3i$ rather than writing the real part of $0$ ).

Adding and Subtracting Complex Numbers

Our goal for the rest of this section is to prove the claim that the complex numbers form a number system just like the real numbers – they can be added, subtracted, multiplied, and divided. We will need some of these skills in the next chapter. In order for this claim to be true, we must be able to write each final answer as a complex number, in the form $a+bi.$

First, consider the following expression.

$(7+6i)+(2+4i)$

If we think of $i$ as a variable for a moment, we just combine the like terms,

$(7+6i)+(2+4i)=9+10i$

The same rules apply to adding and subtracting complex numbers. You combine the corresponding parts and write the final answer again in the form $a+bi.$

Example

Add. $(−3+3i)+(7–2i)$

Show Solution

Rearrange the sums to put like terms together.

$−3+3i+7–2i=−3+7+3i–2i$

Combine like terms.

$−3+7=4$

and

$3i–2i=(3–2)i=i$

The answer is $4+i$ .

Example

Subtract. $(−3+3i)–(7–2i)$

Show Solution

Be sure to distribute the subtraction sign to all terms in the set of parentheses that follows.

$(−3+3i)–(7–2i)=−3+3i–7+2i$

Rearrange the terms to put like terms together.

$−3–7+3i+2i$

Combine like terms.

$−3–7=−10$

and

$3i+2i=(3+2)i=5i$

The answer is $-10+5i$ .

In the following video, we show more examples of how to add and subtract complex numbers (Note in this video that it incorrectly calls $bi$ the imaginary part instead of just $b$ ).

Multiplying Radicals with Negative Radicands

We have to be careful when multiplying radicals with negative radicands. Remember to use the product rule for imaginary numbers first before any other steps.

EXAMPLE

Multiply and simplify $\sqrt{-8}\cdot\sqrt{-12}.$ Write answers in the form $a+bi$

Show Solution

First rewrite each radical using the product rule for imaginary numbers.

$\begin{align}&\quad\sqrt{-8}\cdot\sqrt{-12}\\ =&\quad\sqrt{-1}\cdot\sqrt{8}\cdot\sqrt{-1}\cdot\sqrt{12}\\ =&\quad i\cdot i\cdot\sqrt{8\cdot 12}\\ =&\quad i^2\sqrt{96}\\ =&\quad -\sqrt{96}\end{align}$

The radical can now be simplified normally.

$\begin{align}=&\quad -\sqrt{16}\sqrt{6}\\ =&\quad -4\sqrt{6}\end{align}$

Notice in the previous example, it would be to use our previous product rule for radicals to multiply the radicands: $\sqrt{-8}\cdot\sqrt{-12}\color{red}{\neq \sqrt{-8\cdot -12}=\sqrt{96}}.$ Doing so would cause us to miss the negative sign that should have appeared upon simplifying $i^2.$

Multiplying Complex Numbers

The process of multiplying complex numbers is again similar to the corresponding process for polynomials. The only difference is that we should replace $i^2=-1$ if it appears.

EXAMPLE

Multiply and simplify $3i(-6+5i).$ Write answers in the form $a+bi.$

Show Solution

We can use the distributive property.
$\begin{align}&\quad 3i(-6+5i)\\ =&\quad 3i\cdot (-6) + 3i\cdot 5i&& \color{blue}{\textsf{distribute}}\\ =&\quad -18i+15i^2\\ =&\quad -18i+15(-1) && \color{blue}{\textsf{substitute }i^2=-1}\\ =&\quad -15-18i&& \color{blue}{\textsf{write in the form }a+bi} \end{align}$

Example

Multiply and simplify $\left(4+3i\right)\left(2 - 5i\right).$ Write answers in the form $a+bi.$

Show Solution

Multiply (using FOIL, for example).

\begin{array}{lll} (4 + 3 i) (2 - 5 i) & = (4 \cdot 2) + (4 \cdot (- 5 i)) + (3 i \cdot 2) + ((3 i) \cdot (- 5 i)) \\ = 8 - 20 i + 6 i - 15 i^{2} \\ = 8 - 20 i + 6 i - 15 (- 1) & substitute i^{2} = - 1 \\ = 8 - 20 i + 6 i + 15 \\ = (8 + 15) + (- 20 + 6) i & group like terms together \\ = 23 - 14 i \end{array}

$\begin{array}{lll}\left(4+3i\right)\left(2 - 5i\right) & =\left(4\cdot 2\right)+\left(4\cdot \left(-5i\right)\right)+\left(3i\cdot 2\right)+\left(\left(3i\right)\cdot \left(-5i\right)\right) & \\ & =8-20i+6i-15i^2 & \\ & = 8-20i+6i-15\left(-1\right) & \color{blue}{\textsf{substitute } i^2=-1} \\ & = 8-20i+6i+15 & \\ & = \left(8+15\right)+\left(-20+6\right)i & \color{blue}{\textsf{group like terms together}} \\ & =23-14i & \end{array}$

In the first video, we show more examples of multiplying complex numbers.

Simplifying Powers of $i$

There is a pattern to evaluating powers of $i$ . Let us look at what happens when we raise $i$ to increasing powers.

\begin{aligned} i^{1} & = i \\ i^{2} & = - 1 \\ i^{3} & = i^{2} \cdot i = - 1 \cdot i = - i \\ i^{4} & = i^{3} \cdot i = - i \cdot i = - i^{2} = - (- 1) = 1 \\ i^{5} & = i^{4} \cdot i = 1 \cdot i = i \end{aligned}

$\begin{array}{rl}{i}^{1}&=\quad i\\ {i}^{2}&=\quad -1\\ {i}^{3}&=\quad {i}^{2}\cdot i=-1\cdot i=-i\\ {i}^{4}&=\quad {i}^{3}\cdot i=-i\cdot i=-{i}^{2}=-\left(-1\right)=1\\ {i}^{5}&=\quad {i}^{4}\cdot i=1\cdot i=i\end{array}$

We can see that when we get to the fifth power of

i

$i$ , it is equal to the first power. As we continue to multiply

i

$i$ by itself for increasing powers, we will see a cycle of four. Here are the next four powers of

i

$i$ .

\begin{aligned} i^{6} & = i^{5} \cdot i = i \cdot i = i^{2} = - 1 \\ i^{7} & = i^{6} \cdot i = i^{2} \cdot i = i^{3} = - i \\ i^{8} & = i^{7} \cdot i = i^{3} \cdot i = i^{4} = 1 \\ i^{9} & = i^{8} \cdot i = i^{4} \cdot i = i^{5} = i \end{aligned}

$\begin{array}{rl}{i}^{6}&=\quad {i}^{5}\cdot i=i\cdot i={i}^{2}=-1\\ {i}^{7}&= \quad{i}^{6}\cdot i={i}^{2}\cdot i={i}^{3}=-i\\ {i}^{8}&=\quad {i}^{7}\cdot i={i}^{3}\cdot i={i}^{4}=1\\ {i}^{9}&=\quad {i}^{8}\cdot i={i}^{4}\cdot i={i}^{5}=i\end{array}$

Example

Evaluate ${i}^{35}$ .

Show Solution

Since ${i}^{4}=1$ , we can simplify the problem by factoring out as many factors of ${i}^{4}$ as possible. To do so, first determine how many times $4$ goes into $35$ : $35=4\cdot 8+3$ . The remainder of $3$ is the most important part.

i^{35} = i^{4 \cdot 8 + 3} = i^{4 \cdot 8} \cdot i^{3} = {(i^{4})}^{8} \cdot i^{3} = 1^{8} \cdot i^{3} = i^{3} = - i

${i}^{35}={i}^{4\cdot 8+3}={i}^{4\cdot 8}\cdot {i}^{3}={\left({i}^{4}\right)}^{8}\cdot {i}^{3}={1}^{8}\cdot {i}^{3}={i}^{3}=-i$

The previous example illustrates the easiest way to simplify a large power of $i$ . Rewrite your old exponent as a multiple of $4$ plus a remainder. After some simplifying, the final power of $i$ is just the remainder. Then you just need to know the first four powers of $i.$

In the following video, you will see more examples of how to simplify powers of $i.$

Dividing Complex Numbers

If we write a division of complex numbers in radical form, it should look similar to something we’ve done earlier in this chapter.

$\frac{2+3i}{4-i}=\frac{2+3\sqrt{-1}}{4-\sqrt{-1}}$

We have previously removed radicals from the denominator by rationalizing the denominator using the conjugate. To eliminate the imaginary number $i$ in the denominator, you multiply by the complex conjugate of the denominator which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of $a+bi$ is $a-bi$ .

complex conjugate

The complex conjugate of a complex number $a+bi$ is $a-bi$ . It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. When a complex number is multiplied by its complex conjugate, the result is a real number.

In our previous example, the denominator is $4-i.$ The complex conjugate is $4+i.$ Multiplying, we get $(4-i)(4+i)=16+4i-4i-i^2=16-(-1)=17.$ Let’s use this to finish our division problem.

$\begin{align}&\quad\frac{2+3i}{4-i}\\ =&\quad\frac{2+3i}{4-i}\cdot\frac{4+i}{4+i}\\ =&\quad\frac{(2+3i)(4+i)}{(4-i)(4+i)}\\ =&\quad\frac{8+2i+12i+3i^2}{17}&&\color{blue}{\textsf{FOIL numerator and denominator}}\\ =&\quad\frac{8+2i+12i+3(-1)}{17}\\ =&\quad\frac{5+14i}{17}&&\color{blue}{\textsf{simplify the numerator}}\\ =&\quad\frac{5}{17}+\frac{14}{17}i&&\color{blue}{\textsf{write in the form }a+bi}\end{align}$

We are now done, and most importantly the final answer is in the form of a complex number, $a+bi.$

DIVIDING COMPLEX NUMBERS

Write the division problem as a fraction.
Determine the complex conjugate of the denominator.
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator and simplify.
Write the answer as $a+bi.$

Example

Divide $\left(6-5i\right)$ by $\left(1+3i\right)$ . Write answers in the form $a+bi.$

Show Solution

We begin by writing the problem as a fraction.

\frac{(6 - 5 i)}{(1 + 3 i)}

$\Large\frac{\left(6-5i\right)}{\left(1+3i\right)}$

Then we multiply the numerator and denominator by the complex conjugate of the denominator.

$\begin{array}{rcll}\Large{=\frac{(6-5i)}{(1+3i)}}\cdot\frac{(1-3i)}{(1-3i)}&=&\Large{\frac{6-18i-5i+15{i}^{2}}{1-3i+3i-9{i}^{2}}}& \\&=&\Large{\frac{6-18i-5i+15(-1)}{1-3i+3i-9(-1)}}&\quad\color{blue}{\textsf{substitute }i^2=-1}\\&=&\Large{\frac{-9-23i}{10}}&\quad\color{blue}{\textsf{combine like terms}} \\&=&\Large{-\frac{9}{10}-\frac{23}{10}i}&\quad\color{blue}{\textsf{write in the form }a+bi}\end{array}$

EXAMPLE

Divide $\dfrac{6-8i}{3i}$ . Write answers in the form $a+bi.$

Show Solution

We could multiply the numerator and denominator by the complex conjugate, but since the denominator is a monomial it is easier to just multiply by $-i.$ You will see this still results in the denominator being a real number.

$\begin{align}\frac{6-8i}{3i}&=\frac{\left(6-8i\right)}{3i}\cdot\frac{-i}{-i}\\ &=\frac{-6i+8i^2}{-3i^2}\\ &=\frac{-6i+8(-1)}{-3(-1)}\\ &=\frac{-6i-8}{3}\\ &=\frac{-6i}{3}+\frac{-8}{3}\\ &=-\frac{8}{3}-2i\end{align}$

In the last video, you will see more examples of dividing complex numbers.

Summary

Complex numbers have the form $a+bi$ , where $a$ and $b$ are real numbers and $i$ is the square root of $−1$ . All real numbers can be written as complex numbers by setting $b=0$ . Square roots of negative numbers can be simplified using $\sqrt{-a}=i\sqrt{a}$ where $a>0.$ We can perform arithmetic operations on complex numbers in much the same way as working with polynomials and combining like terms, making sure to simplify $i^2=-1$ when appropriate. Powers of $i$ form a cycle that repeats every four powers. Division of complex numbers is done by first making the denominator a real number using the complex conjugate multiplied on the top and bottom of the fraction.

Module 5: Radical Expressions, Functions, and Equations

Learning Outcomes

product rule for imaginary numbers

Example

Example

Rewriting the Square Root of a Negative Number

Complex Numbers

COMPLEX NUMBERS

Example

Example

Adding and Subtracting Complex Numbers

Example

Example

Multiplying Radicals with Negative Radicands

EXAMPLE

Multiplying Complex Numbers

EXAMPLE

Example

Simplifying Powers of $i$

Example

Dividing Complex Numbers

complex conjugate

DIVIDING COMPLEX NUMBERS

Example

EXAMPLE

Summary

Candela Citations

Module 5: Radical Expressions, Functions, and Equations