2.1: Complex Numbers
- Refresher Topics
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Sets and Numbers
- Sets of Numbers (complex numbers)
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Simplifying Expressions
- Laws of Exponents (Power Law)
- Properties of Radicals and Simplifying Expressions Involving Roots
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Sets and Numbers
In this section, you will:
- Express square roots of negative numbers as multiples of \(i\).
- 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\[x^2 +4=0. \nonumber \]Our best guesses might be \( 2 \) or \( –2 \). But if we test \( 2 \) in this equation, it does not work. If we test \( –2 \), 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 \( i \)
We know how to find the square root of any positive real number. In a similar way, we can find the square root of 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 \(i\) is defined as the square root of negative \( 1 \).\[\sqrt{ −1 } =i. \nonumber \]So, using properties of radicals,\[i^2 = ( \sqrt{ −1 } )^2 =−1. \nonumber \]We can write the square root of any negative number as a multiple of \(i\). Consider the square root of –25.\[\sqrt{ −25 } =\sqrt{ 25 \cdot (−1) } =\sqrt{ 25 } \sqrt{ −1 } =5i. \nonumber \]We use \(5i\) and not \(−5i\) because the principal root of \(25\) 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 \(3+4\sqrt{ 3 }i \).
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.
A complex number is a number of the form \(a+bi\) where
- \(a\) is the real part of the complex number.
- \(b\) is the imaginary part of the complex number.
If \(b=0\), then \(a+bi\) is a real number. If \(a=0\) and \(b\) is not equal to \( 0 \), the complex number is called an imaginary number . An imaginary number is an even root of a negative number. The set symbol for the set of complex numbers is \( \mathbb{C} \).
Express \(\sqrt{ −9 }\) in standard form.
- Solution
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\[\sqrt{ −9 } =\sqrt{ 9 } \sqrt{ −1 } =3i \nonumber \]In standard form, this is \(0+3i\).
Express \(\sqrt{ −24 }\) 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 part and the vertical axis represents the imaginary part. Complex numbers are the points on the plane, expressed as ordered pairs \((a,b)\), where \(a\) represents the coordinate for the horizontal axis and \(b\) represents the coordinate for the vertical axis.
Let's consider the number \(−2+3i\). The real part of the complex number is \(−2\) and the imaginary part is \(3\). We plot the ordered pair \((−2,3)\) to represent the complex number \(−2+3i\) as shown in Figure \( \PageIndex{ 1 } \).
Figure \( \PageIndex{ 1 } \)
In the complex plane , the horizontal axis is the real axis , and the vertical axis is the imaginary axis as shown in Figure \( \PageIndex{ 2 } \).
Figure \( \PageIndex{ 2 } \)
Plot the complex number \(3−4i\) on the complex plane.
- Solution
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The real part of the complex number is \(3\), and the imaginary part is \(−4\). We plot the ordered pair \((3,−4)\) as shown in Figure \( \PageIndex{ 3 } \).
Figure \( \PageIndex{ 3 } \)
Plot the complex number \(−4−i\) on the complex plane.
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.
Adding complex numbers:\[( a+bi )+( c+di )=( a+c )+( b+d )i. \nonumber \]Subtracting complex numbers:\[( a+bi )−( c+di )=( a−c )+( b−d )i. \nonumber \]
Add \(3−4i\) and \(2+5i\).
- Solution
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We add the real parts and add the imaginary parts.\[ \begin{array}{rclcl}
(a + bi) + (c + di) & = & (a + c) + (b + d)i & \quad & \left( \text{regrouping} \right) \\[6pt]
& = & (3 + 2) + (-4 + 5)i & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 5 + i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Subtract \(2+5i\) from \(3–4i\).
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,
Find the product \(4(2+5i)\).
- Solution
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Distribute the 4.\[ \begin{array}{rclcl}
4(2 + 5i) & = & (4 \cdot 2) + (4 \cdot 5i) & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 8 + 20i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Find the product \(−4(2+6i)\).
Multiplying Complex Numbers Together
Now, let's multiply two complex numbers. Using the Distributive Property, we get\[( a+bi )( c+di )=ac+adi+bci+bd i^2. \nonumber \]Because \(i^2 =−1\), we have\[( a+bi )( c+di )=ac+adi+bci−bd. \nonumber \]To simplify, we combine the real parts, and we combine the imaginary parts.\[( a+bi )( c+di )=( ac−bd )+( ad+bc )i. \nonumber \]Rather than memorize this form, it's best to treat complex number multiplication just like you would binomial multiplication.
Multiply \(( 4+3i )(2−5i)\).
- Solution
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\[ \begin{array}{rclcl}
( 4+3i )(2−5i) & = & (4 \cdot 2) - (4 \cdot 5i) + (3i \cdot 2) - (3i \cdot 5i) & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 8 - 20i + 6i - 15 i^2 & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 8 - 14i - 15 (-1) & \quad & \left( \text{combining like terms, and }i^2 = -1 \right) \\[6pt]
& = & 8 - 14i + 15 & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 23 - 14i & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
Multiply \((3−4i)(2+3i)\).
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. This is so important, it bears repeating.
Any fraction must have a real-valued 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 \(a+bi\) is \(a−bi\).
Note that complex conjugates have a reciprocal relationship: The complex conjugate of \(a+bi\) is \(a−bi\), and the complex conjugate of \(a−bi\) is \(a+bi\). 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 \(c+di\) by \(a+bi\), where neither \(a\) nor \(b\) equals zero. We first write the division as a fraction.\[ \dfrac{c+di}{a+bi} \quad \text{where} \quad a \neq 0 \text{ and }b \neq 0. \nonumber \]We then find the complex conjugate of the denominator, which in this case is \( a - bi \), and multiply the numerator and denominator by the complex conjugate of the denominator.\[ \dfrac{(c+di)}{(a+bi)} \cdot \dfrac{( a−bi )}{( a−bi )} = \dfrac{( c+di )( a−bi )}{( a+bi )( a−bi )}. \nonumber \]Applying the distributive property, we get\[\dfrac{ca−cbi+adi−bd i^2}{a^2 −abi+abi− b^2 i^2} = \dfrac{ca−cbi+adi−bd(−1)}{a^2 − b^2 (−1)} = \dfrac{(ca+bd)+(ad−cb)i}{a^2 + b^2}. \nonumber \]As was mentioned with multiplication of complex numbers, it is best not to memorize this resulting formula. You should be able to just "do the math."
Let's codify everything we have just developed.
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.
- When a complex number is added to its complex conjugate, the result is a real number.
- When a complex number is multiplied by its conjugate, the result is always the sum of the square of the real part and the square of the imaginary part.
Find the complex conjugate of each number.
- \(2+i\sqrt{ 5 }\)
- \(−\frac{1}{2} i\)
- Solution
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- The number is already in the form \(a+bi\). The complex conjugate is \(a−bi\), or \(2−i\sqrt{ 5 } \).
- We can rewrite this number in the form \(a+bi\) as \(0− \frac{1}{2} i\). The complex conjugate is \(a−bi\), or \(0+ \frac{1}{2} i\). This can be written simply as \( \frac{1}{2} i\).
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 \(i\).
Divide \(( 2+5i )\) by \(( 4−i )\).
- Solution
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We begin by writing the problem as a fraction:\[ \dfrac{2+5i}{4−i}. \nonumber \]Then we multiply the numerator and denominator by the complex conjugate of the denominator:\[ \dfrac{(2+5i)}{(4−i)} \cdot \dfrac{(4+i)}{(4+i)}. \nonumber \]To multiply two complex numbers, we expand the product as we would with polynomials.\[ \begin{array}{rclcl}
\dfrac{(2+5i)}{(4−i)} \cdot \dfrac{(4+i)}{(4+i)} & = & \dfrac{8 + 2i + 20i + 5i^2}{(4)^2 + (-1)^2} & \quad & \left( \text{distributing and using the fact that }(a + bi)(a - bi) = a^2 + b^2 \right) \\[6pt]
& = & \dfrac{8 + 22i - 5}{16 + 1} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{3 + 22i}{17} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{3}{17} + \dfrac{22}{17}i & \quad & \left( \text{separating the real and imaginary parts} \right) \\[6pt]
\end{array} \nonumber \]Note that this expresses the quotient in standard form.
Let \(f(x)= x^2 −5x+2\). Evaluate \(f( 3+i )\).
- Solution
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\[ \begin{array}{rclcl}
f(3 + i) & = & (3 + i)^2 - 5(3 + i) + 2 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & (9 + 6i + i^2) - 15 - 5 i + 2 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 9 + 6i + (-1) - 15 - 5 i + 2 & \quad & \left( i^2 = -1 \right) \\[6pt]
& = & -5 + i & \quad & \left( \text{combining like terms} \right) \\[6pt]
\end{array} \nonumber \]
Let \(f(x)=2 x^2 −3x\). Evaluate \(f( 8−i )\).
Let \(f( x )= \frac{2+x}{x+3} \). Evaluate \(f( 10i )\).
- Solution
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\[ \begin{array}{rclcl}
f(10i) & = & \dfrac{2 + 10i}{10i + 3} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{2 + 10i}{3 + 10i} & \quad & \left( \text{rewriting the denominator in standard form} \right) \\[6pt]
& = & \dfrac{(2 + 10i)}{(3 + 10i)} \cdot \dfrac{(3 - 10i)}{(3 - 10i)} & \quad & \left( \text{multiplying by the conjugate of the denominator} \right) \\[6pt]
& = & \dfrac{6 - 20i + 30i - 100i^2}{(3)^2 + (10)^2} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{6 - 20i + 30i - 100(-1)}{(3)^2 + (10)^2} & \quad & \left( i^2 = -1 \right) \\[6pt]
& = & \dfrac{106 + 10i}{109} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{106}{109} + \dfrac{10}{109}i & \quad & \left( \text{separating the real and imaginary parts} \right) \\[6pt]
\end{array} \nonumber \]
Let \(f(x)= x+1 x−4 \). Evaluate \(f( −i )\).
Simplifying Powers of \( i \)
The powers of \(i\) are cyclic. Let's look at what happens when we raise \(i\) to increasing powers.\[ \begin{array}{rcl}
i^1 & = & i \\[6pt]
i^2 & = & −1 \\[6pt]
i^3 & = & i^2 \cdot i =−1 \cdot i =−i \\[6pt]
i^4 & = & i^3 \cdot i = −i \cdot i =− i^2 = −(−1) = 1 \\[6pt]
i^5 & = & i^4 \cdot i =1 \cdot i = i \\[6pt]
\end{array} \nonumber \]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 itself for increasing powers, we will see a cycle of 4. Let's examine the next 4 powers of \(i\).\[ \begin{array}{rcl}
i^6 & = & i^5 \cdot i = i \cdot i = i^2 = -1 \\[6pt]
i^7 & = & i^6 \cdot i = i^2 \cdot i = i^3 = -i \\[6pt]
i^8 & = & i^7 \cdot i =i^3 \cdot i = i^4 = 1 \\[6pt]
i^9 & = & i^8 \cdot i = i^4 \cdot i = i^5 = = i \\[6pt]
\end{array} \nonumber \]
Evaluate \(i^{35} \).
- Solution
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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: \(35=4 \cdot 8+3\). Therefore,\[i^{35} = i^{4 \cdot 8+3} = i^{4 \cdot 8} \cdot i^3 = ( i^4 )^8 \cdot i^3 = 1^8 \cdot i^3 = i^3 = −i. \nonumber \]
Can we write \(i^{35}\) in other helpful ways?
As we saw in Example \( \PageIndex{ 10 } \), 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 \(i^{35}\) | \(i^{34} \cdot i\) | \(i^{33} \cdot i^{2}\) | \(i^{31} \cdot i^4\) | \(i^{19} \cdot i^{16}\) |
| Reduced form | \(( i^2 )^{17} \cdot i\) | \(i^{33} \cdot ( −1 )\) | \(i^{31} \cdot 1\) | \(i^{19} \cdot ( i^4 )^4\) |
| Simplified form | \(( −1 )^{17} \cdot i\) | \(− i^{33}\) | \(i^{31}\) | \(i^{19}\) |
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
Access these online resources for additional instruction and practice with complex numbers.