2.1: Complex Numbers
- Page ID
- 193551
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Add and subtract complex numbers.
- Multiply and divide complex numbers.
- Solve quadratic equations with complex numbers
Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.
To better understand it, we need to become familiar with a new set of numbers. Keep in mind that 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 fills a void left by the set of integers. The set of irrational numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it.
Expressing Square Roots of Negative Numbers as Multiples of \(i\)
We know how to find the square root of any positive real number. In a similar way, we can find the square root of any 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 or complex number.The imaginary number \(i\) is defined as the square root of \(−1\).
\[\sqrt{-1}=i\]
Using properties of radicals,
\[i^2=(\sqrt{-1})^2=-1\]
We can write the square root of any negative number as a multiple of \(i\). Consider the square root of \(−49\).
\[\begin{align*} \sqrt{-49}&= \sqrt{49\times(-1)}\\[4pt] &= \sqrt{49}\sqrt{-1}\\[4pt] &= 7i \end{align*}\]
We use \(7i\) and not \(−7i\) because the principal root of \(49\) 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 \(a+bi\) where \(a\) is the real part and \(b\) is the imaginary part. For example, \(5+2i\) is a complex number. So, too, is \(3+4i\sqrt{3}\).
Imaginary numbers differ from real numbers in that 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, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.
a. Express \(\sqrt{-9}\) in standard form.
Solution
\[\begin{align*} \sqrt{-9}&= \sqrt{9}\sqrt{-1)}\\[4pt] &= 3i\\[4pt] \end{align*}\]
In standard form, this is \(0+3i\).
b. Express \(\sqrt{-24}\) in standard form.
Solution
\[\begin{align*} \sqrt{-24}&= \sqrt{24}\sqrt{-1)}\\[4pt] &= 2\sqrt{6}\sqrt{-1}&=2i\sqrt{6}\\[4pt] \end{align*}\]
In standard form, this is \(0+2i\sqrt{6}\)
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 then combine the imaginary parts.
Adding complex numbers:
\[(a+bi)+(c+di)=(a+c)+(b+d)i)\]
Subtracting complex numbers:
\[(a+bi)−(c+di)=(a−c)+(b−d)i\]
Add or subtract as indicated.
- \((3−4i)+(2+5i)\)
- \((−5+7i)−(−11+2i)\)
Solution
- \[\begin{align*} (3-4i)+(2+5i)&= 3-4i+2+5i\\[4pt] &= 3+2+(-4i)+5i\\[4pt] &= (3+2)+(-4+5)i\\[4pt] &= 5+i \end{align*}\]
- \[\begin{align*} (-5+7i)-(-11+2i)&= -5+7i+11-2i\\[4pt] &= -5+11+7i-2i\\[4pt] &= (-5+11)+(7-2)i\\[4pt] &= 6+5i \end{align*}\]
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
Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, \(3(6+2i)\) :
Find the product \(4(2+5i)\).
Solution
Distribute the \(4\).
\[\begin{align*} 4(2+5i)&= (4\cdot 2)+(4\cdot 5i)\\[4pt] &= 8+20i \end{align*}\]
Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, \(i^2\), it equals \(-1\).
\[\begin{align*} (a+bi)(c+di)&= ac+adi+bci+bdi^2\\[4pt] &= ac+adi+bci-bd(-1)\qquad i^2 = -1\\[4pt] &= ac+adi+bci-bd\\[4pt] &= (ac-bd)+(ad+bc)i \end{align*}\]
Group real terms and imaginary terms.
Multiply \((4+3i)(2−5i)\).
Solution
\[\begin{align*} (4+3i)(2-5i)&= 4(2)-4(5i)+3i(2)-(3i)(5i)\\[4pt] &= 8-20i+6i-15(i^2)\\[4pt] &= (8+15)+(-20+6)i\\[4pt] &= 23-14i \end{align*}\]
Simplifying Powers of \(i\)
The powers of \(i\) are cyclic. Let’s look at what happens when we raise \(i\) to increasing powers.
\[i^1=i \nonumber \] \[i^2=-1 \nonumber \] \[i^3=i^2⋅i=-1⋅i=-i \nonumber \] \[i^4=i^3⋅i=-i⋅i=-i^2=-(-1)=1 \nonumber \] \[i^5=i^4⋅i=1⋅i=i \nonumber \]
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 increasing powers, we will see a cycle of four. Let’s examine the next four powers of \(i\).
\[i^6=i^5⋅i=i⋅i=i^2=-1 \nonumber \] \[i^7=i^6⋅i=i^2⋅i=i^3=-i \nonumber \] \[i^8=i^7⋅i=i^3⋅i=i^4=1 \nonumber \] \[i^9=i^8⋅i=i^4⋅i=i^5=i \nonumber \]
The cycle is repeated continuously: \(i,−1,−i,1,\) every four powers.
Evaluate: \(i^{35}\).
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 \(35: 35=4⋅8+3\).
\[i^{35}=i^{4⋅8+3}=i^{4⋅8}⋅i^3={(i^4)}^8⋅i^3=i^8⋅i^3=i^3=−i \nonumber \]
Can we write \(i^{35}\) in other helpful ways?
As we saw in Example \(\PageIndex{8}\), we reduced \(i^{35}\) to \(i^3\) by dividing the exponent by \(4\) and using the remainder to find the simplified form. But perhaps another factorization of \(i^{35}\) may be more useful. Table \(\PageIndex{1}\) shows some other possible factorizations.
| Factorization of \(i^{35}\) | \(i^{34}⋅i\) | \(i^{33}⋅i^2\) | \(i^{31}⋅i^4\) | \(i^{19}⋅i^{16}\) |
|---|---|---|---|---|
| Reduced form | \({(i^2)}^{17}⋅i\) | \(i^{33}⋅(−1)\) | \(i^{31}⋅1\) | \(i^{19}⋅{(i^4)}^4\) |
| Simplified form | \({(−1)}^{17}⋅i\) | \(−i^{33}\) | \(i^{31}\) | \(i^{19}\) |
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
Key Concepts
- The square root of any negative number can be written as a multiple of \(i\).
- Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example.
- Complex numbers can be multiplied and divided.
- To multiply complex numbers, distribute just as with polynomials.
- The powers of i are cyclic, repeating every fourth one.