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Use the Complex Number System

Use the Complex Number System

By the end of this section, you will be able to: evaluate the square root of a negative number, add and subtract complex numbers, multiply complex numbers, divide complex numbers, simplify powers of ii.

Evaluate the Square Root of a Negative Number

Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify −1,\sqrt{-1}, we are looking for a real number xx so that x2=−1x^2=-1. Since all real numbers squared are positive numbers, there is no real number that equals −1-1 when squared.

Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like 5.\sqrt{5}. All of these together gave us the real numbers and so far in your study of mathematics, that has been sufficient.

But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit ii as the number whose square is −1-1.

Imaginary Unit

The imaginary unit ii is the number whose square is −1-1.

i2=−1 or i=−1i^{2} = -1\ \text{or}\ i = \sqrt{-1}

We will use the imaginary unit to simplify the square roots of negative numbers.

Square Root of a Negative Number

If b is a positive real number, then

−b=bi\sqrt{- b} = \sqrt{b} i

We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as −b=ib\sqrt{- b} = i \sqrt{b} to emphasize the i is not under the radical. But the −b=bi\sqrt{- b} = \sqrt{b} i is considered standard form.

Example.

Write each expression in terms of i and simplify if possible:

(a) −25\sqrt{-25} (b) −7\sqrt{-7} (c) −12.\sqrt{-12}.

(a)

StepResult
−25\sqrt{-25}
Use the definition of the square root of negative numbers.25i\sqrt{25} i
Simplify.5i5 i

(b)

StepResult
−7\sqrt{-7}
Use the definition of the square root of negative numbers.7i\sqrt{7} i
Simplify.Be careful that it is clear that ii is not under the radical sign.

(c)

StepResult
−12\sqrt{-12}
Use the definition of the square root of negative numbers.12i\sqrt{12} i
Simplify 12.\sqrt{12}.23i2 \sqrt{3} i

Write each expression in terms of ii and simplify if possible: −81\sqrt{-81}

Write each expression in terms of ii and simplify if possible: −5\sqrt{-5}

Write each expression in terms of ii and simplify if possible: −18\sqrt{-18}

Now that we are familiar with the imaginary number i, we can expand our concept of the number system to include imaginary numbers. The complex number system includes the real numbers and the imaginary numbers. A complex number is of the form a + bi, where a, b are real numbers. We call a the real part and b the imaginary part.

Complex Number

A complex number is of the form a + bi, where a and b are real numbers.

a⏟real part+bi⏟imaginary term \underbrace{a}_{\text{real part}}+\underbrace{bi}_{\text{imaginary term}}

A complex number is in standard form when written as a+bi,a + b i, where a and b are real numbers.

If b=0,b = 0, then a+bia + b i becomes a+0⋅i=a,a + 0 \cdot i = a, and is a real number.

If b≠0,b \neq 0, then a+bia + b i is an imaginary number.

If a=0,a = 0, then a+bia + b i becomes 0+bi=bi,0 + b i = b i, and is called a pure imaginary number.

We summarize this here.

StepResultResult
a+bia + b i
b=0b = 0a+0⋅ia\begin{matrix} \\ a + 0 \cdot i \\ \\ a\end{matrix}Real number
b≠0b \neq 0a+bia + b iImaginary number
a=0a = 00+bibi\begin{matrix}0 + b i \\ \\ \\ b i\end{matrix}Pure imaginary number

The standard form of a complex number is a+bi,a + b i, so this explains why the preferred form is −b=bi\sqrt{- b} = \sqrt{b} i when b>0.b > 0.

The table helps us visualize the complex number system. It is made up of both the real numbers and the imaginary numbers.

Add or Subtract Complex Numbers

We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.

Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in standard form.

Example.

Add: −12+−27.\sqrt{-12} + \sqrt{-27}.

StepResult
−12+−27\sqrt{-12} + \sqrt{-27}
Use the definition of the square root of negative numbers.12i+27i\sqrt{12} i + \sqrt{27} i
Simplify the square roots.23i+33i2 \sqrt{3} i + 3 \sqrt{3} i
Add.53i5 \sqrt{3} i

Add: −8+−32\sqrt{-8} + \sqrt{-32}

Add: −27+−48\sqrt{-27} + \sqrt{-48}

Remember to add both the real parts and the imaginary parts in this next example.

Example.

Simplify: (a) (4−3i)+(5+6i)\left(4 - 3 i\right) + \left(5 + 6 i\right) (b) (2−5i)−(5−2i).\left(2 - 5 i\right) - \left(5 - 2 i\right).

(a)

StepResult
(4−3i)+(5+6i)\left(4 - 3 i\right) + \left(5 + 6 i\right)
Use the Associative Property to put the realparts and the imaginary parts together.(4+5)+(−3i+6i)\left(4 + 5\right) + \left(-3 i + 6 i\right)
Simplify.9+3i9 + 3 i

(b)

StepResult
(2−5i)−(5−2i)\left(2 - 5 i\right) - \left(5 - 2 i\right)
Distribute.2−5i−5+2i2 - 5 i - 5 + 2 i
Use the Associative Property to put the realparts and the imaginary parts together.2−5−5i+2i2 - 5 - 5 i + 2 i
Simplify.−3−3i-3 - 3 i

Simplify: (2+7i)+(4−2i)\left(2 + 7 i\right) + \left(4 - 2 i\right)

Simplify: (8−4i)−(2−i)\left(8 - 4 i\right) - \left(2 - i\right)

Simplify: (3−2i)+(−5−4i)\left(3 - 2 i\right) + \left(-5 - 4 i\right)

Multiply Complex Numbers

Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in the next two examples.

Example.

Multiply: 2i(7−5i).2 i \left(7 - 5 i\right).

StepResult
2i(7−5i)2 i \left(7 - 5 i\right)
Distribute.14i−10i214 i - 10 i^{2}
Simplify i2.i^{2}.14i−10(−1)14 i - 10 \left(-1\right)
Multiply.14i+1014 i + 10
Write in standard form.10+14i10 + 14 i

Multiply: 4i(5−3i)4 i \left(5 - 3 i\right)

Multiply: −3i(2+4i)-3 i \left(2 + 4 i\right)

In the next example, we multiply the binomials using the Distributive Property or FOIL.

Example.

Multiply: (3+2i)(4−3i).\left(3 + 2 i\right) \left(4 - 3 i\right).

StepResult
(3+2i)(4−3i)\left(3 + 2 i\right) \left(4 - 3 i\right)
Use FOIL.12−9i+8i−6i212 - 9 i + 8 i - 6 i^{2}
Simplify i2i^{2} and combine like terms.12−i−6(−1)12 - i - 6 \left(-1\right)
Multiply.12−i+612 - i + 6
Combine the real parts.18−i18 - i

Multiply: (5−3i)(−1−2i)\left(5 - 3 i\right) \left(-1 - 2 i\right)

Multiply: (−4−3i)(2+i)\left(-4 - 3 i\right) \left(2 + i\right)

In the next example, we could use FOIL or the Product of Binomial Squares Pattern.

Example.

Multiply: (3+2i)2\left(3 + 2 i\right)^{2}

StepResult
(3+2i)2(3+2i)^2
Use the Product of Binomial Squares Pattern, (a+b)2=a2+2ab+b2.\left(a + b\right)^{2} = a^{2} + 2 a b + b^{2}.32+2(3)(2i)+(2i)23^2+2(3)(2i)+(2i)^2
Simplify.9+12i+4i29+12i+4i^2
Simplify i2.i^{2}.9+12i+4(−1)9+12i+4(-1)
Simplify.5+12i5+12i

Multiply using the Binomial Squares pattern: (−2−5i)2\left(-2 - 5 i\right)^{2}

Multiply using the Binomial Squares pattern: (−5+4i)2\left(-5 + 4 i\right)^{2}

Since the square root of a negative number is not a real number, when we have the square roots of two negative numbers, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as complex numbers, using −b=bi.\sqrt{- b} = \sqrt{b} i. This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Example.

Multiply: −36⋅−4.\sqrt{-36} \cdot \sqrt{-4}.

To multiply square roots of negative numbers, we first write them as complex numbers.

StepResult
−36⋅−4\sqrt{-36} \cdot \sqrt{-4}
Write as complex numbers using −b=bi.\sqrt{- b} = \sqrt{b} i.36i⋅4i\sqrt{36} i \cdot \sqrt{4} i
Simplify.6i⋅2i6 i \cdot 2 i
Multiply.12i212 i^{2}
Simplify i2i^{2} and multiply.−12-12

Multiply: −49⋅−4\sqrt{-49} \cdot \sqrt{-4}

Multiply: −36⋅−81\sqrt{-36} \cdot \sqrt{-81}

In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.

Example.

Multiply: (3−−12)(5+−27).\left(3 - \sqrt{-12}\right) \left(5 + \sqrt{-27}\right).

To multiply square roots of negative numbers, we first write them as complex numbers.

StepResult
(3−−12)(5+−27)\left(3 - \sqrt{-12}\right) \left(5 + \sqrt{-27}\right)
Write as complex numbers using −b=bi.\sqrt{- b} = \sqrt{b} i.(3−23i)(5+33i)\left(3 - 2 \sqrt{3} i\right) \left(5 + 3 \sqrt{3} i\right)
Use FOIL.15+93i−103i−6⋅3i215 + 9 \sqrt{3} i - 10 \sqrt{3} i - 6 \cdot 3 i^{2}
Combine like terms and simplify i2.i^{2}.15−3i−6⋅(−3)15 - \sqrt{3} i - 6 \cdot \left(-3\right)
Multiply and combine like terms.33−3i33 - \sqrt{3} i

Multiply: (4−−12)(3−−48)\left(4 - \sqrt{-12}\right) \left(3 - \sqrt{-48}\right)

Multiply: (−2+−8)(3−−18)\left(-2 + \sqrt{-8}\right) \left(3 - \sqrt{-18}\right)

We first looked at conjugate pairs when we studied polynomials. We said that a pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form (a−b),(a+b).\left(a - b\right), \left(a + b\right).

A complex conjugate pair is very similar. For a complex number of the form a+bi,a + b i, its conjugate is a−bi.a - b i. Notice they have the same first term and the same last term, but one is a sum and one is a difference.

Complex Conjugate Pair

A complex conjugate pair is of the form a+bi,a + b i, a−bi.a - b i.

We will multiply a complex conjugate pair in the next example.

Example.

Multiply: (3−2i)(3+2i).\left(3 - 2 i\right) \left(3 + 2 i\right).

StepResult
(3−2i)(3+2i)\left(3 - 2 i\right) \left(3 + 2 i\right)
Use FOIL.9+6i−6i−4i29 + 6 i - 6 i - 4 i^{2}
Combine like terms and simplify i2.i^{2}.9−4(−1)9 - 4 \left(-1\right)
Multiply and combine like terms.13

Multiply: (4−3i)⋅(4+3i)\left(4 - 3 i\right) \cdot \left(4 + 3 i\right)

Multiply: (−2+5i)⋅(−2−5i)\left(-2 + 5 i\right) \cdot \left(-2 - 5 i\right)

From our study of polynomials, we know the product of conjugates is always of the form (a−b)(a+b)=a2−b2.\left(a - b\right) \left(a + b\right) = a^{2} - b^{2}. The result is called a difference of squares. We can multiply a complex conjugate pair using this pattern.

The last example used FOIL. Now we use the Product of Conjugates Pattern:

(3−2i)(3+2i)=32−(2i)2=9−4i2=9−4(−1)=13. \begin{aligned} (3-2i)(3+2i) &=3^2-(2i)^2 \\ &=9-4i^2 \\ &=9-4(-1) \\ &=13. \end{aligned}

Notice this is the same result we found in Example 8.84.

When we multiply complex conjugates, the product of the last terms will always have an i2i^{2} which simplifies to −1.-1.

(a−bi)(a+bi)a2−(bi)2a2−b2i2a2−b2(−1)a2+b2\begin{matrix}\left(a - b i\right) \left(a + b i\right) \\ a^{2} - \left(b i\right)^{2} \\ a^{2} - b^{2} i^{2} \\ a^{2} - b^{2} \left(-1\right) \\ a^{2} + b^{2}\end{matrix}

This leads us to the Product of Complex Conjugates Pattern: (a−bi)(a+bi)=a2+b2\left(a - b i\right) \left(a + b i\right) = a^{2} + b^{2}

Product of Complex Conjugates

If a and b are real numbers, then

(a−bi)(a+bi)=a2+b2\left(a - b i\right) \left(a + b i\right) = a^{2} + b^{2}

Example.

Multiply using the Product of Complex Conjugates Pattern: (8−2i)(8+2i).\left(8 - 2 i\right) \left(8 + 2 i\right).

StepResult
(8−2i)(8+2i)(8-2i)(8+2i)
Use the Product of Complex Conjugates Pattern, (a−bi)(a+bi)=a2+b2.\left(a - b i\right) \left(a + b i\right) = a^{2} + b^{2}.82+228^2+2^2
Simplify the squares.64+464+4
Add.6868

Multiply using the Product of Complex Conjugates Pattern: (3−10i)(3+10i)\left(3 - 10 i\right) \left(3 + 10 i\right)

Multiply using the Product of Complex Conjugates Pattern: (−5+4i)(−5−4i)\left(-5 + 4 i\right) \left(-5 - 4 i\right)

Divide Complex Numbers

Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.

Example.

How to Divide Complex Numbers

Divide: 4+3i3−4i.\tfrac{4 + 3 i}{3 - 4 i}.

StepResult
1. Write the numerator and denominator in standard form. They already are.4+3i3−4i\tfrac{4+3i}{3-4i}
2. Multiply the numerator and denominator by the complex conjugate of the denominator.4+3i3−4i⋅3+4i3+4i\tfrac{4+3i}{3-4i}\cdot\tfrac{3+4i}{3+4i}
Multiply.(4+3i)(3+4i)(3−4i)(3+4i)\tfrac{(4+3i)(3+4i)}{(3-4i)(3+4i)}
3. Simplify the numerator and use the Product of Complex Conjugates Pattern in the denominator.12+16i+9i+12i29+16\tfrac{12+16i+9i+12i^2}{9+16}
Combine like terms and use i2=−1i^2=-1.12+25i−1225\tfrac{12+25i-12}{25}
Simplify and write the result in standard form.25i25=i\tfrac{25i}{25}=i

Divide: 2+5i5−2i\tfrac{2 + 5 i}{5 - 2 i}

Divide: 1+6i6−i\tfrac{1 + 6 i}{6 - i}

We summarize the steps here.

How To

How to divide complex numbers.

  1. Step 1. Write both the numerator and denominator in standard form.
  2. Step 2. Multiply the numerator and denominator by the complex conjugate of the denominator.
  3. Step 3. Simplify and write the result in standard form.

Example.

Divide, writing the answer in standard form: −35+2i.\tfrac{-3}{5 + 2 i}.

StepResult
−35+2i\frac{-3}{5 + 2 i}
Multiply the numerator and denominator by thecomplex conjugate of the denominator.−3(5−2i)(5+2i)(5−2i)\frac{-3 \left(5 - 2 i\right)}{\left(5 + 2 i\right) \left(5 - 2 i\right)}
Multiply in the numerator and use the Product ofComplex Conjugates Pattern in the denominator.−15+6i52+22\frac{-15 + 6 i}{5^{2} + 2^{2}}
Simplify.−15+6i29\frac{-15 + 6 i}{29}
Write in standard form.−1529+629i- \frac{15}{29} + \frac{6}{29} i

Divide, writing the answer in standard form: 41−4i\tfrac{4}{1 - 4 i}

Divide, writing the answer in standard form: −2−1+2i\tfrac{-2}{-1 + 2 i}

Be careful as you find the conjugate of the denominator.

Example.

Divide: 5+3i4i.\tfrac{5 + 3 i}{4 i}.

StepResult
5+3i4i\frac{5 + 3 i}{4 i}
Write the denominator in standard form.5+3i0+4i\frac{5 + 3 i}{0 + 4 i}
Multiply the numerator and denominator by the complex conjugate of the denominator.(5+3i)(0−4i)(0+4i)(0−4i)\frac{\left(5 + 3 i\right) \left(0 - 4 i\right)}{\left(0 + 4 i\right) \left(0 - 4 i\right)}
Simplify.(5+3i)(−4i)(4i)(−4i)\frac{\left(5 + 3 i\right) \left(-4 i\right)}{\left(4 i\right) \left(-4 i\right)}
Multiply.−20i−12i2−16i2\frac{-20 i - 12 i^{2}}{-16 i^{2}}
Simplify the i2.i^{2}.−20i+1216\frac{-20 i + 12}{16}
Rewrite in standard form.1216−2016i\frac{12}{16} - \frac{20}{16} i
Simplify the fractions.34−54i\frac{3}{4} - \frac{5}{4} i

Divide: 3+3i2i\tfrac{3 + 3 i}{2 i}

Divide: 2+4i5i\tfrac{2 + 4 i}{5 i}

Simplify Powers of i

The powers of ii make an interesting pattern that will help us simplify higher powers of i. Let’s evaluate the powers of ii to see the pattern.

i1i2i3i4i−1i2⋅ii2⋅i2−1⋅i(−1)(−1)−i1i5i6i7i8i4⋅ii4⋅i2i4⋅i3i4⋅i41⋅i1⋅i21⋅i31⋅1ii2i31−1−i\begin{matrix}i^{1} & & & i^{2} & & & i^{3} & & & i^{4} \\ i & & & - 1 & & & i^{2} \cdot i & & & i^{2} \cdot i^{2} \\ & & & & & & - 1 \cdot i & & & \left(-1\right) \left(-1\right) \\ & & & & & & - i & & & 1 \\ \\ \\ i^{5} & & & i^{6} & & & i^{7} & & & i^{8} \\ i^{4} \cdot i & & & i^{4} \cdot i^{2} & & & i^{4} \cdot i^{3} & & & i^{4} \cdot i^{4} \\ 1 \cdot i & & & 1 \cdot i^{2} & & & 1 \cdot i^{3} & & & 1 \cdot 1 \\ i & & & i^{2} & & & i^{3} & & & 1 \\ & & & - 1 & & & - i\end{matrix}

We summarize this now.

i1=ii5=ii2=−1i6=−1i3=−ii7=−ii4=1i8=1\begin{aligned}i^{1} & = & i & & & i^{5} & = & i \\ i^{2} & = & -1 & & & i^{6} & = & -1 \\ i^{3} & = & - i & & & i^{7} & = & - i \\ i^{4} & = & 1 & & & i^{8} & = & 1\end{aligned}

If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of ii. Since i4=1i^4=1, we rewrite each power, ini^n, as a product using i4i^4 to a power and another power of ii.

We rewrite it in the form in=(i4)q⋅ir,i^{n} = \left(i^{4}\right)^{q} \cdot i^{r}, where the exponent, qq, is the quotient of nn divided by 4 and the exponent, rr, is the remainder from this division. For example, to simplify i57i^{57}, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, 57=4⋅14+1.57 = 4 \cdot 14 + 1. Therefore,

i57=(i4)14i1=114i=i. i^{57}=\left(i^4\right)^{14}i^1=1^{14}i=i.

Example.

Simplify: i86.i^{86}.

StepResult
i86i^{86}
Divide 86 by 4. The quotient is 21 and the remainder is 2.86=4(21)+286=4(21)+2
Rewrite i86i^{86} in the in=(i4)q⋅iri^{n} = \left(i^{4}\right)^{q} \cdot i^{r} form.(i4)21⋅i2\left(i^{4}\right)^{21} \cdot i^{2}
Simplify.(1)21⋅(−1)\left(1\right)^{21} \cdot \left(-1\right)
Simplify.−1-1

Simplify: i75i^{75}

Simplify: i92i^{92}

This section is adapted from Intermediate Algebra 2e, Section 8.8: Use the Complex Number System by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted worked solutions for the web; omitted the Be Prepared quiz, media links, self-check reflection, and end-of-section exercise bank; and converted the source Try It practice into interactive exercises.

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