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Complex Numbers

By the end of this section, you will be able to:

  • Express square roots of negative numbers as multiples of ii
  • Plot complex numbers on the complex plane
  • Add and subtract complex numbers
  • Multiply and divide complex numbers

The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as

x2+4=0x^2+4=0

Our best guesses might be +2+2 or 2-2. But if we test +2+2 in this equation, it does not work. If we test 2-2, it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it.

Expressing square roots of negative numbers as multiples of ii

We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number ii is defined as the square root of negative 1.

1=i\sqrt{-1}=i

So, using properties of radicals,

i2=(1)2=1i^2=\left(\sqrt{-1}\right)^2=-1

We can write the square root of any negative number as a multiple of ii. Consider the square root of 25-25.

25=25(1)=251=5i \begin{array}{lrcl} & \sqrt{-25} &=& \sqrt{25\cdot(-1)} \\[4pt] & &=& \sqrt{25}\sqrt{-1} \\[4pt] & &=& 5i \end{array}

We use 5i5i and not 5i-5i because the principal root of 2525 is the positive root.

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+bia+bi, where aa is the real part and bibi is the imaginary part. For example, 5+2i5+2i is a complex number, in which 55 is the real part and 2i2i is the imaginary part. So, too, is 3+43i3+4\sqrt{3}i.

Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall that 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.

Imaginary and complex numbers. A complex number is a number of the form a+bia+bi where

  • aa is the real part of the complex number.
  • bibi is the imaginary part of the complex number.

If b=0b=0, then a+bia+bi is a real number. If a=0a=0 and bb is not equal to 00, the complex number is called an imaginary number. An imaginary number is an even root of a negative number.

How to: given an imaginary number, express it in standard form.

  1. Write a\sqrt{-a} as a1\sqrt{a}\sqrt{-1}.
  2. Express 1\sqrt{-1} as ii.
  3. Write ai\sqrt{a}\cdot i in simplest form.

Example. Express 9\sqrt{-9} in standard form.

Solution.

9=91=3i\sqrt{-9}=\sqrt{9}\sqrt{-1}=3i

In standard form, this is 0+3i0+3i.

Express24\sqrt{-24}in standard form.

Plotting a complex number on the complex plane

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a,b)(a,b), where aa represents the coordinate for the horizontal axis and bb represents the coordinate for the vertical axis.

Let’s consider the number 2+3i-2+3i. The real part of the complex number is 2-2 and the imaginary part is 3i3i. We plot the ordered pair (2,3)(-2,3) to represent the complex number 2+3i-2+3i as shown below.

Complex plane. In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown below.

How to: given a complex number, represent its components on the complex plane.

  1. Determine the real part and the imaginary part of the complex number.
  2. Move along the horizontal axis to show the real part of the number.
  3. Move parallel to the vertical axis to show the imaginary part of the number.
  4. Plot the point.

Example. Plot the complex number 34i3-4i on the complex plane.

Solution. The real part of the complex number is 33, and the imaginary part is 4i-4i. We plot the ordered pair (3,4)(3,-4) as shown below.

Plot the complex number4i-4-ion the complex plane by giving its ordered pair(a,b)(a,b).

Adding and subtracting complex numbers

Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.

Complex numbers: addition and subtraction. Adding complex numbers:

(a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)+(c+di)=(a+c)+(b+d)i

Subtracting complex numbers:

(a+bi)(c+di)=(ac)+(bd)i(a+bi)-(c+di)=(a-c)+(b-d)i

How to: given two complex numbers, find the sum or difference.

  1. Identify the real and imaginary parts of each number.
  2. Add or subtract the real parts.
  3. Add or subtract the imaginary parts.

Example. Add 34i3-4i and 2+5i2+5i.

Solution. We add the real parts and add the imaginary parts.

(a+bi)+(c+di)=(a+c)+(b+d)i(34i)+(2+5i)=(3+2)+(4+5)i=5+i \begin{array}{lrcl} & (a+bi)+(c+di) &=& (a+c)+(b+d)i \\[4pt] & (3-4i)+(2+5i) &=& (3+2)+(-4+5)i \\[4pt] & &=& 5+i \end{array}

Subtract2+5i2+5ifrom34i3-4i.

Multiplying 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

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 example,

3(6+2i)=(36)+(32i)=18+6i3(6+2i)=(3\cdot6)+(3\cdot2i)=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 4(2+5i)4(2+5i).

Solution. Distribute the 4.

4(2+5i)=(42)+(45i)=8+20i \begin{array}{lrcl} & 4(2+5i) &=& (4\cdot2)+(4\cdot5i) \\[4pt] & &=& 8+20i \end{array}

Find the product4(2+6i)-4(2+6i).

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

(a+bi)(c+di)=ac+adi+bci+bdi2(a+bi)(c+di)=ac+adi+bci+bdi^2

Because i2=1i^2=-1, we have

(a+bi)(c+di)=ac+adi+bcibd(a+bi)(c+di)=ac+adi+bci-bd

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

(a+bi)(c+di)=(acbd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)i

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

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

Example. Multiply (4+3i)(25i)(4+3i)(2-5i).

Solution. Use (a+bi)(c+di)=(acbd)+(ad+bc)i(a+bi)(c+di)=(ac-bd)+(ad+bc)i.

(4+3i)(25i)=(423(5))+(4(5)+32)i=(8+15)+(20+6)i=2314i \begin{array}{lrcl} & (4+3i)(2-5i) &=& \left(4\cdot2-3\cdot(-5)\right)+\left(4\cdot(-5)+3\cdot2\right)i \\[4pt] & &=& (8+15)+(-20+6)i \\[4pt] & &=& 23-14i \end{array}

Multiply(34i)(2+3i)(3-4i)(2+3i).

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. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called 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+bia+bi is abia-bi.

Note that complex conjugates have a reciprocal relationship: the complex conjugate of a+bia+bi is abia-bi, and the complex conjugate of abia-bi is a+bia+bi. 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 c+dic+di by a+bia+bi, where neither aa nor bb equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

c+dia+bi where a0 and b0\tfrac{c+di}{a+bi}\ \text{where}\ a\ne0\ \text{and}\ b\ne0

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

(c+di)(a+bi)(abi)(abi)=(c+di)(abi)(a+bi)(abi)\tfrac{(c+di)}{(a+bi)}\cdot\tfrac{(a-bi)}{(a-bi)}=\tfrac{(c+di)(a-bi)}{(a+bi)(a-bi)}

Apply the distributive property.

cacbi+adibdi2a2abi+abib2i2\tfrac{ca-cbi+adi-bdi^2}{a^2-abi+abi-b^2i^2}

Simplify, remembering that i2=1i^2=-1.

=cacbi+adibd(1)a2abi+abib2(1)=(ca+bd)+(adcb)ia2+b2 \begin{array}{lrcl} & &=& \tfrac{ca-cbi+adi-bd(-1)}{a^2-abi+abi-b^2(-1)} \\[4pt] & &=& \tfrac{(ca+bd)+(ad-cb)i}{a^2+b^2} \end{array}

The complex conjugate. The complex conjugate of a complex number a+bia+bi is abia-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.
  • When a complex number is added to its complex conjugate, the result is a real number.

Example. Find the complex conjugate of each number.

(a) 2+i52+i\sqrt{5}

(b) 12i-\tfrac{1}{2}i

Solution.

(a) The number is already in the form a+bia+bi. The complex conjugate is abia-bi, or 2i52-i\sqrt{5}.

(b) We can rewrite this number in the form a+bia+bi as 012i0-\tfrac{1}{2}i. The complex conjugate is abia-bi, or 0+12i0+\tfrac{1}{2}i. This can be written simply as 12i\tfrac{1}{2}i.

Analysis. 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 ii.

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 (2+5i)(2+5i) by (4i)(4-i).

Solution. We begin by writing the problem as a fraction.

(2+5i)(4i)\tfrac{(2+5i)}{(4-i)}

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

(2+5i)(4i)(4+i)(4+i)\tfrac{(2+5i)}{(4-i)}\cdot\tfrac{(4+i)}{(4+i)}

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

(2+5i)(4i)(4+i)(4+i)=8+2i+20i+5i216+4i4ii2Because i2=1.=8+2i+20i+5(1)16+4i4i(1)=3+22i17Separate real and imaginary parts.=317+2217i \begin{array}{lrcl} & \tfrac{(2+5i)}{(4-i)}\cdot\tfrac{(4+i)}{(4+i)} &=& \tfrac{8+2i+20i+5i^2}{16+4i-4i-i^2} \\[4pt] \text{Because}\ i^2=-1. & &=& \tfrac{8+2i+20i+5(-1)}{16+4i-4i-(-1)} \\[4pt] & &=& \tfrac{3+22i}{17} \\[4pt] \text{Separate real and imaginary parts.} & &=& \tfrac{3}{17}+\tfrac{22}{17}i \end{array}

Note that this expresses the quotient in standard form.

Example. Let f(x)=x25x+2f(x)=x^2-5x+2. Evaluate f(3+i)f(3+i).

Solution. Substitute x=3+ix=3+i into the function f(x)=x25x+2f(x)=x^2-5x+2 and simplify.

f(3+i)=(3+i)25(3+i)+2Multiply.=(3+6i+i2)(15+5i)+2Substitute 1 for i2.=9+6i+(1)155i+2Combine like terms.=5+i \begin{array}{lrcl} & f(3+i) &=& (3+i)^2-5(3+i)+2 \\[4pt] \text{Multiply.} & &=& (3+6i+i^2)-(15+5i)+2 \\[4pt] \text{Substitute}\ -1\ \text{for}\ i^2. & &=& 9+6i+(-1)-15-5i+2 \\[4pt] \text{Combine like terms.} & &=& -5+i \end{array}

Analysis. We write f(3+i)=5+if(3+i)=-5+i. Notice that the input is 3+i3+i and the output is 5+i-5+i.

Letf(x)=2x23xf(x)=2x^2-3x. Evaluatef(8i)f(8-i).

Example. Let f(x)=2+xx+3f(x)=\tfrac{2+x}{x+3}. Evaluate f(10i)f(10i).

Solution. Substitute x=10ix=10i and simplify.

f(10i)=2+10i10i+3Rewrite the denominator in standard form.=2+10i3+10iMultiply by the complex conjugate of the denominator.=2+10i3+10i310i310iMultiply using the distributive property.=620i+30i100i2930i+30i100i2Substitute 1 for i2.=620i+30i100(1)930i+30i100(1)Simplify.=106+10i109Separate the real and imaginary parts.=106109+10109i \begin{array}{lrcl} & f(10i) &=& \tfrac{2+10i}{10i+3} \\[4pt] \text{Rewrite the denominator in standard form.} & &=& \tfrac{2+10i}{3+10i} \\[4pt] \text{Multiply by the complex conjugate of the denominator.} & &=& \tfrac{2+10i}{3+10i}\cdot\tfrac{3-10i}{3-10i} \\[4pt] \text{Multiply using the distributive property.} & &=& \tfrac{6-20i+30i-100i^2}{9-30i+30i-100i^2} \\[4pt] \text{Substitute}\ -1\ \text{for}\ i^2. & &=& \tfrac{6-20i+30i-100(-1)}{9-30i+30i-100(-1)} \\[4pt] \text{Simplify.} & &=& \tfrac{106+10i}{109} \\[4pt] \text{Separate the real and imaginary parts.} & &=& \tfrac{106}{109}+\tfrac{10}{109}i \end{array}

Letf(x)=x+1x4f(x)=\tfrac{x+1}{x-4}. Evaluatef(i)f(-i).

Simplifying powers of ii

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

i1=ii2=1i3=i2i=1i=ii4=i3i=ii=i2=(1)=1i5=i4i=1i=i \begin{array}{lrcl} & i^1 &=& i \\[4pt] & i^2 &=& -1 \\[4pt] & i^3 &=& i^2\cdot i=-1\cdot i=-i \\[4pt] & i^4 &=& i^3\cdot i=-i\cdot i=-i^2=-(-1)=1 \\[4pt] & i^5 &=& i^4\cdot i=1\cdot i=i \end{array}

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

i6=i5i=ii=i2=1i7=i6i=i2i=i3=ii8=i7i=i3i=i4=1i9=i8i=i4i=i5=i \begin{array}{lrcl} & i^6 &=& i^5\cdot i=i\cdot i=i^2=-1 \\[4pt] & i^7 &=& i^6\cdot i=i^2\cdot i=i^3=-i \\[4pt] & i^8 &=& i^7\cdot i=i^3\cdot i=i^4=1 \\[4pt] & i^9 &=& i^8\cdot i=i^4\cdot i=i^5=i \end{array}

Example. Evaluate i35i^{35}.

Solution. Since i4=1i^4=1, we can simplify the problem by factoring out as many factors of i4i^4 as possible. To do so, first determine how many times 4 goes into 35: 35=48+335=4\cdot8+3.

i35=i48+3=i48i3=(i4)8i3=18i3=i3=ii^{35}=i^{4\cdot8+3}=i^{4\cdot8}\cdot i^3=\left(i^4\right)^8\cdot i^3=1^8\cdot i^3=i^3=-i

Q&A. Can we write i35i^{35} in other helpful ways?

As shown above, we reduced i35i^{35} to i3i^3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35i^{35} may be more useful. The table below shows some other possible factorizations.

Factorization of i35i^{35}i34ii^{34}\cdot ii33i2i^{33}\cdot i^2i31i4i^{31}\cdot i^4i19i16i^{19}\cdot i^{16}
Reduced form(i2)17i\left(i^2\right)^{17}\cdot ii33(1)i^{33}\cdot(-1)i311i^{31}\cdot1i19(i4)4i^{19}\cdot\left(i^4\right)^4
Simplified form(1)17i(-1)^{17}\cdot ii33-i^{33}i31i^{31}i19i^{19}

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

Media. Access these online resources for additional instruction and practice with complex numbers.

Key concepts

  • The square root of any negative number can be written as a multiple of ii.
  • To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.
  • Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts.
  • Complex numbers can be multiplied and divided.
  • To multiply complex numbers, distribute just as with polynomials.
  • To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator.
  • The powers of ii are cyclic, repeating every fourth one.

Practice

Express square roots of negative numbers as multiples of ii

Perform the indicated operation and express the result as a simplified complex number:9+316\sqrt{-9}+3\sqrt{-16}.

Perform the indicated operation and express the result as a simplified complex number:2+122\tfrac{2+\sqrt{-12}}{2}.

Plot complex numbers on the complex plane

Give the ordered pair(a,b)(a,b)used to plot the complex number12i1-2ion the complex plane.

Give the ordered pair(a,b)(a,b)used to plot the complex numberiion the complex plane.

Add and subtract complex numbers

Perform the indicated operation and express the result as a simplified complex number:(3+2i)+(53i)(3+2i)+(5-3i).

Perform the indicated operation and express the result as a simplified complex number:(5+3i)(6i)(-5+3i)-(6-i).

Perform the indicated operation and express the result as a simplified complex number:(4+4i)(6+9i)(-4+4i)-(-6+9i).

Multiply and divide complex numbers

Perform the indicated operation and express the result as a simplified complex number:(52i)(3i)(5-2i)(3i).

Perform the indicated operation and express the result as a simplified complex number:(1+2i)(2+3i)(-1+2i)(-2+3i).

Perform the indicated operation and express the result as a simplified complex number:(3+4i)(34i)(3+4i)(3-4i).

Perform the indicated operation and express the result as a simplified complex number:3+4i2i\tfrac{3+4i}{2-i}.

Perform the indicated operation and express the result as a simplified complex number:6+4ii\tfrac{6+4i}{i}.

Iff(x)=x2+3x+5f(x)=x^2+3x+5, evaluatef(2+i)f(2+i).


This section is adapted from Precalculus 2e, Section 3.1: Complex Numbers by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the section’s three complex-plane figures as accessible inline SVG built from the exact plotted point — 2+3i-2+3i, the bare real/imaginary axis pair, and 34i3-4i — each drawn as a horizontal move along the real axis followed by a vertical move to the labelled point, matching the source’s two-arrow diagrams; omitted the decorative “5+2i5+2i real part / imaginary part” callout artwork and the “3(6+2i)3(6+2i)” distribution-arrows illustration, folding their content into the surrounding prose and a plain step equation instead; reconstructed the worked step-by-step for evaluating f(3+i)f(3+i), where f(x)=x25x+2f(x)=x^2-5x+2, from the source’s page image, since its CNXML solution is an image with no transcribed steps, and independently verified the result 5+i-5+i by substitution; omitted the two “count the real and nonreal solutions from a graphed parabola” exercises, whose source art has no transcribable geometry; omitted the “Access these online resources” media links, keeping only the introductory sentence; converted the section’s “Try It” checks into interactive components, including two complex-plane plotting checks rewritten as ordered-pair fill-ins (GraphPlot only grades a line, a system of two lines, or a quadratic, not a plotted point); and adapted 13 selected end-of-section exercises — two radical-to-ii simplifications, two complex-plane plotting conversions, three addition/subtraction simplifications, two multiplications, a product of complex conjugates, two divisions (one by a complex denominator, one by ii) written in standard a+bia+bi form, and a polynomial evaluated at a complex input — into interactive components in a closing Practice block, one group per objective. Every complex-division answer in this section, in both the exposition and the Practice block, is authored in standard a+bia+bi form rather than as a single fraction over a complex denominator, because the pinned compute-engine build computes complex division incorrectly when the denominator itself is complex.