We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > College Algebra: Co-requisite Course

Imaginary and Complex Numbers

Learning Objectives

  • Define imaginary and complex numbers
    • Express roots of negative numbers in terms of i.
    • Express imaginary numbers as bi and complex numbers as [latex]a+bi[/latex].
  • Algebraic operations on complex numbers
    • Add complex numbers.
    • Subtract complex numbers
    • Multiply complex numbers.
    • Find conjugates of complex numbers.
    • Divide complex numbers.
    • Simplify powers of i
Up to now, you’ve known it was impossible to take a square root of a negative number. This is true, using only the real numbers. But here you will learn about a new kind of number that lets you work with square roots of negative numbers! Complex numbers are made from both real and imaginary numbers.  Imaginary numbers are called imaginary because they are impossible and, therefore, exist only in the world of ideas and pure imagination. Imaginary numbers result from taking the square root of a negative number. Here we will first define and perform algebraic operations on complex numbers, then we will provide examples of quadratic equations that have solutions that are complex numbers. You really need only one new number to start working with the square roots of negative numbers. That number is the square root of [latex]−1,\sqrt{-1}[/latex]. The real numbers are those that can be shown on a number line—they seem pretty real to us! When something’s not real, you often say it is imaginary. So let’s call this new number i and use it to represent the square root of [latex]−1[/latex].

[latex] i=\sqrt{-1}[/latex]

Because [latex] \sqrt{x}\,\cdot \,\sqrt{x}=x[/latex], we can also see that [latex] \sqrt{-1}\,\cdot \,\sqrt{-1}=-1[/latex] or [latex] i\,\cdot \,i=-1[/latex]. We also know that [latex] i\,\cdot \,i={{i}^{2}}[/latex], so we can conclude that [latex] {{i}^{2}}=-1[/latex].

[latex] {{i}^{2}}=-1[/latex]

The number i allows us to work with roots of all negative numbers, not just [latex] \sqrt{-1}[/latex]. There are two important rules to remember: [latex] \sqrt{-1}=i[/latex], and [latex] \sqrt{ab}=\sqrt{a}\sqrt{b}[/latex]. You will use these rules to rewrite the square root of a negative number as the square root of a positive number times [latex] \sqrt{-1}[/latex]. Next you will simplify the square root and rewrite [latex] \sqrt{-1}[/latex] as i. Let’s try an example.

Example

Simplify. [latex] \sqrt{-4}[/latex]

Answer: Use the rule [latex] \sqrt{ab}=\sqrt{a}\sqrt{b}[/latex] to rewrite this as a product using [latex] \sqrt{-1}[/latex]. [latex-display] \sqrt{-4}=\sqrt{4\cdot -1}=\sqrt{4}\sqrt{-1}[/latex-display] Since 4 is a perfect square [latex](4=2^{2})[/latex], you can simplify the square root of 4. [latex-display] \sqrt{4}\sqrt{-1}=2\sqrt{-1}[/latex-display] Use the definition of i to rewrite [latex] \sqrt{-1}[/latex] as i. [latex-display] 2\sqrt{-1}=2i[/latex-display]

Answer

[latex-display] \sqrt{-4}=2i[/latex-display]

Example

Simplify. [latex] \sqrt{-18}[/latex]

Answer: Use the rule [latex] \sqrt{ab}=\sqrt{a}\sqrt{b}[/latex] to rewrite this as a product using [latex] \sqrt{-1}[/latex]. [latex-display] \sqrt{-18}=\sqrt{18\cdot -1}=\sqrt{18}\sqrt{-1}[/latex-display] Since 18 is not a perfect square, use the same rule to rewrite it using factors that are perfect squares. In this case, 9 is the only perfect square factor, and the square root of 9 is 3. [latex-display] \sqrt{18}\sqrt{-1}=\sqrt{9}\sqrt{2}\sqrt{-1}=3\sqrt{2}\sqrt{-1}[/latex-display] Use the definition of i to rewrite [latex] \sqrt{-1}[/latex] as i. [latex-display] 3\sqrt{2}\sqrt{-1}=3\sqrt{2}i=3i\sqrt{2}[/latex-display] Remember to write i in front of the radical.

Answer

[latex-display] \sqrt{18}=3i\sqrt[{}]{2}[/latex-display]

Example

Simplify. [latex] -\sqrt{-72}[/latex]

Answer: Use the rule [latex] \sqrt{ab}=\sqrt{a}\sqrt{b}[/latex] to rewrite this as a product using [latex] \sqrt{-1}[/latex]. [latex-display] -\sqrt{-72}=-\sqrt{72\cdot -1}=-\sqrt{72}\sqrt{-1}[/latex-display] Since 72 is not a perfect square, use the same rule to rewrite it using factors that are perfect squares. Notice that 72 has three perfect squares as factors: 4, 9, and 36. It’s easiest to use the largest factor that is a perfect square. [latex-display] -\sqrt{72}\sqrt{-1}=-\sqrt{36}\sqrt{2}\sqrt{-1}=-6\sqrt{2}\sqrt{-1}[/latex-display] Use the definition of i to rewrite [latex] \sqrt{-1}[/latex] as i. [latex-display] -6\sqrt{2}\sqrt{-1}=-6\sqrt{2}i=-6i\sqrt{2}[/latex-display] Remember to write i in front of the radical.

Answer

[latex-display] -\sqrt{-}72=-6i\sqrt[{}]{2}[/latex-display]

You may have wanted to simplify [latex] -\sqrt{-72}[/latex] using different factors. Some may have thought of rewriting this radical as [latex] -\sqrt{-9}\sqrt{8}[/latex], or [latex] -\sqrt{-4}\sqrt{18}[/latex], or [latex] -\sqrt{-6}\sqrt{12}[/latex] for instance. Each of these radicals would have eventually yielded the same answer of [latex] -6i\sqrt{2}[/latex]. In the following video, we show more examples of how to use imaginary numbers to simplify a square root with a negative radicand. https://youtu.be/LSp7yNP6Xxc

Rewriting the Square Root of a Negative Number

  • Find perfect squares within the radical.
  • Rewrite the radical using the rule [latex] \sqrt{ab}=\sqrt{a}\cdot \sqrt{b}[/latex].
  • Rewrite [latex] \sqrt{-1}[/latex] as i.
Example: [latex] \sqrt{-18}=\sqrt{9}\sqrt{-2}=\sqrt{9}\sqrt{2}\sqrt{-1}=3i\sqrt{2}[/latex]

Complex Numbers

Showing the real and imaginary parts of 5 + 2i. In this complex number, 5 is the real part and 2i is the complex part.

A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part. For example, [latex]5+2i[/latex] is a complex number. So, too, is [latex]3+4\sqrt{3}i[/latex].

Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.

You can use the usual operations (addition, subtraction, multiplication, and so on) with imaginary numbers. You’ll see more of that, later. When you add a real number to an imaginary number, however, you get a complex number. A complex number is any number in the form [latex]a+bi[/latex], where a is a real number and bi is an imaginary number. The number a is sometimes called the real part of the complex number, and bi is sometimes called the imaginary part.
Complex Number Real part Imaginary part
[latex]3+7i[/latex] 3 [latex]7i[/latex]
[latex]18–32i[/latex] 18 [latex]−32i[/latex]
[latex] -\frac{3}{5}+i\sqrt{2}[/latex] [latex] -\frac{3}{5}[/latex] [latex] i\sqrt{2}[/latex]
[latex] \frac{\sqrt{2}}{2}-\frac{1}{2}i[/latex] [latex] \frac{\sqrt{2}}{2}[/latex] [latex]-\frac{1}{2}i[/latex]
In a number with a radical as part of b, such as [latex]-\frac{3}{5}+i\sqrt{2}[/latex] above, the imaginary i should be written in front of the radical. Though writing this number as [latex] -\frac{3}{5}+\sqrt{2}i[/latex] is technically correct, it makes it much more difficult to tell whether i is inside or outside of the radical. Putting it before the radical, as in [latex] -\frac{3}{5}+i\sqrt{2}[/latex], clears up any confusion. Look at these last two examples.
Number Number in complex form: [latex]a+bi[/latex] Real part Imaginary part
17 [latex]17+0i[/latex] 17 [latex]0i[/latex]
[latex]−3i[/latex] [latex]0–3i[/latex] 0 [latex]−3i[/latex]
By making [latex]b=0[/latex], any real number can be expressed as a complex number. The real number a is written [latex]a+0i[/latex] in complex form. Similarly, any imaginary number can be expressed as a complex number. By making [latex]a=0[/latex], any imaginary number [latex]bi[/latex] is written [latex]0+bi[/latex] in complex form.

Example

Write 83.6 as a complex number.

Answer: Remember that a complex number has the form [latex]a+bi[/latex]. You need to figure out what a and b need to be. [latex-display]a+bi[/latex-display] Since 83.6 is a real number, it is the real part (a) of the complex number [latex]a+bi[/latex]. A real number does not contain any imaginary parts, so the value of b is 0. [latex-display]83.6+bi[/latex-display]

Answer

[latex-display]83.6+0i[/latex-display]

Example

Write [latex]−3i[/latex] as a complex number.

Answer: Remember that a complex number has the form [latex]a+bi[/latex]. You need to figure out what a and b need to be. [latex-display]a+bi[/latex-display] Since [latex]−3i[/latex] is an imaginary number, it is the imaginary part (bi) of the complex number [latex]a+bi[/latex]. This imaginary number has no real parts, so the value of a is 0. [latex-display]a–3i[/latex-display]

Answer

[latex-display]0–3i[/latex-display]

In the next video we show more examples of how to write numbers as complex numbers. https://youtu.be/mfoOYdDkuyY

Add and Subtract Complex Numbers

Any time new kinds of numbers are introduced, one of the first questions that needs to be addressed is, “How do you add them?” In this topic, you’ll learn how to add complex numbers and also how to subtract. First, consider the following expression.

[latex](6x+8)+(4x+2)[/latex]

To simplify this expression, you combine the like terms, [latex]6x[/latex] and [latex]4x[/latex]. These are like terms because they have the same variable with the same exponents. Similarly, 8 and 2 are like terms because they are both constants, with no variables.

[latex](6x+8)+(4x+2)=10x+10[/latex]

In the same way, you can simplify expressions with radicals.

[latex] (6\sqrt{3}+8)+(4\sqrt{3}+2)=10\sqrt{3}+10[/latex]

You can add [latex] 6\sqrt{3}[/latex] to [latex] 4\sqrt{3}[/latex] because the two terms have the same radical, [latex] \sqrt{3}[/latex], just as 6x and 4x have the same variable and exponent. The number i looks like a variable, but remember that it is equal to [latex]\sqrt{-1}[/latex]. The great thing is you have no new rules to worry about—whether you treat it as a variable or a radical, the exact same rules apply to adding and subtracting complex numbers. You combine the imaginary parts (the terms with i), and you combine the real parts.

Example

Add. [latex](−3+3i)+(7–2i)[/latex]

Answer: Rearrange the sums to put like terms together.

[latex]−3+3i+7–2i=−3+7+3i–2i[/latex]

Combine like terms.

[latex]−3+7=4[/latex] and [latex]3i–2i=(3–2)i=i[/latex]

Answer

[latex-display](−3+3i)+(7–2i)=4+i[/latex-display]

Example

Subtract. [latex](−3+3i)–(7–2i)[/latex]

Answer: Be sure to distribute the subtraction sign to all terms in the subtrahend.

[latex](−3+3i)–(7–2i)=−3+3i–7+2i[/latex] 

Rearrange the terms to put like terms together.

[latex]−3–7+3i+2i[/latex]

Combine like terms.

[latex]−3–7=−10[/latex] and [latex]3i+2i=(3+2)i=5i[/latex]

Answer

[latex-display](−3+3i)–(7–2i)=10+5i[/latex-display]

In the following video we show more examples of how to add and subtract complex numbers. https://youtu.be/SGhTjioGqqA

Multiply and Divide Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

 Multiplying a Complex Number by a Real Number

Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i. Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So,for 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.

How To: Given a complex number and a real number, multiply to find the product.

  1. Use the distributive property.
  2. Simplify.

Example

Find the product [latex]4\left(2+5i\right)[/latex].

Answer:

Distribute the 4.

[latex]\begin{array}{cc}4\left(2+5i\right)=\left(4\cdot 2\right)+\left(4\cdot 5i\right)\hfill \\ =8+20i\hfill \end{array}[/latex]

 Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci+bd{i}^{2}[/latex]

Because [latex]{i}^{2}=-1[/latex], we have

[latex]\left(a+bi\right)\left(c+di\right)=ac+adi+bci-bd[/latex]

To simplify, we combine the real parts, and we combine the imaginary parts.

[latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]

How To: Given two complex numbers, multiply to find the product.

  1. Use the distributive property or the FOIL method.
  2. Simplify.

Example

Multiply [latex]\left(4+3i\right)\left(2 - 5i\right)[/latex].

Answer:

Use [latex]\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i[/latex]

[latex]\begin{array}{ccc}\left(4+3i\right)\left(2 - 5i\right)=\left(4\cdot 2 - 3\cdot \left(-5\right)\right)+\left(4\cdot \left(-5\right)+3\cdot 2\right)i\hfill \\ \text{ }=\left(8+15\right)+\left(-20+6\right)i\hfill \\ \text{ }=23 - 14i\hfill \end{array}[/latex]

In the first video we show more examples of multiplying complex numbers. https://youtu.be/Fmr3o2zkwLM

Simplifying Powers of i

The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.

[latex]\begin{array}{cc}{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}=-\left(-1\right)=1\\ {i}^{5}={i}^{4}\cdot i=1\cdot i=i\end{array}[/latex]

We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.

[latex]\begin{array}{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{array}[/latex]

Example

Evaluate [latex]{i}^{35}[/latex].

Answer:

Since [latex]{i}^{4}=1[/latex], we can simplify the problem by factoring out as many factors of [latex]{i}^{4}[/latex] as possible. To do so, first determine how many times 4 goes into 35: [latex]35=4\cdot 8+3[/latex].

[latex]{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[/latex]

Q & A

Can we write [latex]{i}^{35}[/latex] in other helpful ways?

As we saw in Example 11, we reduced [latex]{i}^{35}[/latex] to [latex]{i}^{3}[/latex] by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of [latex]{i}^{35}[/latex] may be more useful. The table below shows some other possible factorizations.

Factorization of [latex]{i}^{35}[/latex] [latex]{i}^{34}\cdot i[/latex] [latex]{i}^{33}\cdot {i}^{2}[/latex] [latex]{i}^{31}\cdot {i}^{4}[/latex] [latex]{i}^{19}\cdot {i}^{16}[/latex]
Reduced form [latex]{\left({i}^{2}\right)}^{17}\cdot i[/latex] [latex]{i}^{33}\cdot \left(-1\right)[/latex] [latex]{i}^{31}\cdot 1[/latex] [latex]{i}^{19}\cdot {\left({i}^{4}\right)}^{4}[/latex]
Simplified form [latex]{\left(-1\right)}^{17}\cdot i[/latex] [latex]-{i}^{33}[/latex] [latex]{i}^{31}[/latex] [latex]{i}^{19}[/latex]

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

In the following video you will see more examples of how to simplify powers of i. https://youtu.be/sfP6SmEYHRw

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. This idea is similar to rationalizing the denominator of a fraction that contains a radical. To eliminate the complex or imaginary number 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 [latex]a+bi[/latex] is [latex]a-bi[/latex].

Note that complex conjugates have a reciprocal relationship: The complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex], and the complex conjugate of [latex]a-bi[/latex] is [latex]a+bi[/latex]. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide [latex]c+di[/latex] by [latex]a+bi[/latex], where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

[latex]\displaystyle \frac{c+di}{a+bi}\text{ where }a\ne 0\text{ and }b\ne 0[/latex]

Multiply the numerator and denominator by the complex conjugate of the denominator.

[latex]\displaystyle \frac{\left(c+di\right)}{\left(a+bi\right)}\cdot \frac{\left(a-bi\right)}{\left(a-bi\right)}=\frac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}[/latex]

Apply the distributive property.

[latex]\displaystyle =\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}[/latex]

Simplify, remembering that [latex]{i}^{2}=-1[/latex].

[latex]\begin{array}=\frac{ca-cbi+adi-bd\left(-1\right)}{{a}^{2}-abi+abi-{b}^{2}\left(-1\right)}\hfill \\ =\frac{\left(ca+bd\right)+\left(ad-cb\right)i}{{a}^{2}+{b}^{2}}\hfill \end{array}[/latex]

A General Note: The Complex Conjugate

The complex conjugate of a complex number [latex]a+bi[/latex] is [latex]a-bi[/latex]. 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.
  • When a complex number is added to its complex conjugate, the result is a real number.

Example

Find the complex conjugate of each number.

  1. [latex]2+i\sqrt{5}[/latex]
  2. [latex]-\frac{1}{2}i[/latex]

Answer:

  1. The number is already in the form [latex]a+bi//[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]2-i\sqrt{5}[/latex].
  2. We can rewrite this number in the form [latex]a+bi[/latex] as [latex]0-\frac{1}{2}i[/latex]. The complex conjugate is [latex]a-bi[/latex], or [latex]0+\frac{1}{2}i[/latex]. This can be written simply as [latex]\frac{1}{2}i[/latex].

Analysis of the Solution

Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.

In the last video you will see more examples of dividing complex numbers.
https://youtu.be/XBJjbJAwM1c

How To: Given two complex numbers, divide one by the other.

  1. Write the division problem as a fraction.
  2. Determine the complex conjugate of the denominator.
  3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
  4. Simplify.

Example

Divide [latex]\left(2+5i\right)[/latex] by [latex]\left(4-i\right)[/latex].

Answer:

We begin by writing the problem as a fraction.

[latex]\displaystyle \frac{\left(2+5i\right)}{\left(4-i\right)}[/latex]

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

[latex]\displaystyle \frac{\left(2+5i\right)}{\left(4-i\right)}\cdot \frac{\left(4+i\right)}{\left(4+i\right)}[/latex]

To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).

[latex]\begin{array}\frac{\left(2+5i\right)}{\left(4-i\right)}\cdot \frac{\left(4+i\right)}{\left(4+i\right)}=\frac{8+2i+20i+5{i}^{2}}{16+4i - 4i-{i}^{2}}\hfill & \hfill \\ \text{ }=\frac{8+2i+20i+5\left(-1\right)}{16+4i - 4i-\left(-1\right)}\hfill & \text{Because } {i}^{2}=-1\hfill \\ \text{ }=\frac{3+22i}{17}\hfill & \hfill \\ \text{ }=\frac{3}{17}+\frac{22}{17}i\hfill & \text{Separate real and imaginary parts}.\hfill \end{array}[/latex]

Note that this expresses the quotient in standard form.

Licenses & Attributions

CC licensed content, Original

  • Write Number in the Form of Complex Numbers. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Simplify Square Roots to Imaginary Numbers. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.

CC licensed content, Shared previously

  • Ex 1: Adding and Subtracting Complex Numbers. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download fro free at: http://cnx.org/contents/[email protected]:1/Preface.
  • Ex: Raising the imaginary unit i to powers. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex: Dividing Complex Numbers. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.