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6.7: Complex Numbers

  • Page ID
    66358
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    Learning Objectives

    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 \(i\)
    Be Prepared

    Before you get started, take this readiness quiz.

    1. Given the numbers \(-4,-\sqrt{7}, 0 . \overline{5}, \dfrac{7}{3}, 3, \sqrt{81}\), list the
      1. rational numbers
      2. irrational numbers
      3. real numbers
    2. Multiply: \((x−3)(2x+5)\).
    3. Rationalize the denominator: \(\dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{3}}\)

    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 \(\sqrt{-1}\), we are looking for a real number \(x\) so that \(x^{2}=-1\). Since all real numbers squared are positive numbers, there is no real number that equals \(–1\) 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 \(\sqrt{5}\). 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 \(i\) as the number whose square is \(–1\).

    Definition \(\PageIndex{1}\)

    The imaginary unit \(i\) is the number whose square is \(-1\).\

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

    Definition \(\PageIndex{2}\)

    Square Root of a Negative Number

    If \(b\) is a positive real number, then

    \(\sqrt{-b}=\sqrt{b} i.\)

    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 \(\sqrt{-b}=i \sqrt{b}\) to emphasize the \(i\) is not under the radical. But the \(\sqrt{-b}=\sqrt{b} i\) is considered standard form.

    Example \(\PageIndex{3}\)

    Write each expression in terms of \(i\) and simplify is possible:

    1. \(\sqrt{-25}\)
    2. \(\sqrt{-7}\)
    3. \(\sqrt{-12}\)

    Solution:

    a.

    \(\sqrt{-25}\)

    Use the definition of the square root of negative numbers.

    \(\sqrt{25} i\)

    Simplify.

    \(5i\)

    b.

    \(\sqrt{-7}\)

    Use the definition of the square root of negative numbers.

    \(\sqrt{7} i\)

    Simplify.

    Be careful that it is clear that \(i\) is not under the radical sign.

    c.

    \(\sqrt{-12}\)

    Use the definition of the square root of negative numbers.

    \(\sqrt{12} i\)

    Simplify \(\sqrt{12}\).

    \(2 \sqrt{3} i\)

    Try It \(\PageIndex{4}\)

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

    1. \(\sqrt{-81}\)
    2. \(\sqrt{-5}\)
    3. \(\sqrt{-18}\)
    Answer
    1. \(9i\)
    2. \(\sqrt{5} i\)
    3. \(3 \sqrt{2} i\)
    Try It \(\PageIndex{5}\)

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

    1. \(\sqrt{-36}\)
    2. \(\sqrt{-3}\)
    3. \(\sqrt{-27}\)
    Answer
    1. \(6i\)
    2. \(\sqrt{3} i\)
    3. \(3\sqrt{3} i\)

    Now that we are familiar with the imaginary number \(i\), we can expand the real numbers 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.

    Definition \(\PageIndex{6}\) 

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

    The image shows the expression a plus b i. The number a is labeled “real part” and the number b i is labeled “imaginary part”.
     

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

    If \(b=0\), then \(a+bi\) becomes \(a+0⋅i=a\), and is a real number.

    If \(b≠0\), then \(a+bi\) is an imaginary number.

    If \(a=0\), then \(a+bi\) becomes \(0+bi=bi\), and is called a pure imaginary number.

    We summarize this here.

      \(a+bi\)  
    \(b=0\)

    \(a+0 \cdot i\)

    \(a\)

    Real number
    \(b\neq 0\) \(a+bi\) Imaginary number
    \(a=0\)R

    \(0+bi\)

    \(bi\)

    Pure imaginary numbe4

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

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

    The table has four rows and three columns. The first row is a header and the second column entry a plus b i. In the second row is b equals zero, a plus 0 i, and “Real number”. The third row contains b is not equal to 0, a plus b i, and “Imaginary number”. The fourth row contains a = 0, 0 plus b i, and “Pure imaginary number”.
    Figure 8.8.2

    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 \(\PageIndex{7}\)

    Add: \(\sqrt{-12}+\sqrt{-27}\).

    Solution:

    \(\sqrt{-12}+\sqrt{-27}\)

    Use the definition of the square root of negative numbers.

    \(\sqrt{12} i+\sqrt{27} i\)

    Simplify the square roots.

    \(2 \sqrt{3} i+3 \sqrt{3} i\)

    Add.

    \(5 \sqrt{3} i\)

    Try It \(\PageIndex{8}\)

    Add: \(\sqrt{-8}+\sqrt{-32}\).

    Answer

    \(6 \sqrt{2} i\)

    Try It \(\PageIndex{9}\)

    Add: \(\sqrt{-27}+\sqrt{-48}\).

    Answer

    \(7 \sqrt{3} i\)

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

    Example \(\PageIndex{10}\)

    Simplify:

    1. \((4-3 i)+(5+6 i)\)
    2. \((2-5 i)-(5-2 i)\)

    Solution:

    a.

    Use the Associative Property to put the real parts and the imaginary parts together.

    Simplify.

    \(9+3i\)

    b.

    Distribute.

    Use the Associative Property to put the real parts and the imaginary parts together.

    Simplify.

    Try It \(\PageIndex{11}\)

    Simplify:

    1. \((2+7 i)+(4-2 i)\)
    2. \((8-4 i)-(2-i)\)
    Answer
    1. \(6+5i\)
    2. \(6-3i\)
    Try It \(\PageIndex{12}\)

    Simplify:

    1. \((3-2 i)+(-5-4 i)\)
    2. \((4+3 i)-(2-6 i)\)
    Answer
    1. \(-2-6i\)
    2. \(2+9i\)

    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 \(\PageIndex{13}\)

    Multiply: \(2 i(7-5 i)\).

    Solution:

    Distribute.

    Simplify \(i^{2}\).

    \(14 i-10(-1)\)

    Multiply.

    \(14 i+10\)

    Write in standard form.

    \(10+14i\)

    Try It \(\PageIndex{14}\)

    Multiply: \(4 i(5-3 i)\).

    Answer

    \(12+20i\)

    Try It \(\PageIndex{15}\)

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

    Answer

    \(12+6i\)

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

    Example \(\PageIndex{16}\)

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

    Solution:

    Use FOIL.

    Simplify \(i^{2}\) and combine like terms.

    \(12-i-6(-1)\)

    Multiply.

    \(12-i+6\)

    Combine the real parts.

    \(18-i\)

    Try It \(\PageIndex{17}\)

    Multiple: \((5-3 i)(-1-2 i)\).

    Answer

    \(-11-7i\)

    Try It \(\PageIndex{18}\)

    Multiple: \((-4-3 i)(2+i)\).

    Answer

    \(-5-10i\)

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

    Example \(\PageIndex{19}\)

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

    Solution:

      .
    Use the Product of Binomial Squares Pattern, \((a+b)^{2}=a^{2}+2 a b+b^{2}\). .
    Simplify. .
    Simplify \(i^{2}\). .
    Simplify. .
     
    Try It \(\PageIndex{20}\)

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

    Answer

    \(-21-20 i\)

    Try It \(\PageIndex{21}\)

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

    Answer

    \(9-40i\)

    Since the square root of a negative number is not a real number, 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 \(\sqrt{-b}=\sqrt{b}i\).This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

    Example \(\PageIndex{22}\)

    Multiply: \(\sqrt{-36} \cdot \sqrt{-4}\).

    Solution:

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

    \(\sqrt{-36} \cdot \sqrt{-4}\)

    Write as complex numbers using \(\sqrt{-b}=\sqrt{b}i\).

    \(\sqrt{36} i \cdot \sqrt{4} i\)

    Simplify.

    \(6 i \cdot 2 i\)

    Multiply.

    \(12i^{2}\)

    Simplify \(i^{2}\) and multiply.

    \(-12\)

    Try It \(\PageIndex{23}\)

    Multiply: \(\sqrt{-49} \cdot \sqrt{-4}\).

    Answer

    \(-14\)

    Try It \(\PageIndex{24}\)

    Multiply: \(\sqrt{-36} \cdot \sqrt{-81}\).

    Answer

    \(-54\)

    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 \(\PageIndex{25}\)

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

    Solution:

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

    \((3-\sqrt{-12})(5+\sqrt{-27})\)

    Write as complex numbers using \(\sqrt{-b}=\sqrt{b}i\).

    Use FOIL.

    Combine like terms and simplify \(i^{2}\).

    Multiply and combine like terms.

    \(33-\sqrt{3} i\)

    Try It \(\PageIndex{26}\)

    Multiply: \((4-\sqrt{-12})(3-\sqrt{-48})\).

    Answer

    \(-12-22 \sqrt{3} i\)

    Try It \(\PageIndex{27}\)

    Multiply: \((-2+\sqrt{-8})(3-\sqrt{-18})\).

    Answer

    \(6+12 \sqrt{2} i\)

    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 \((a−b),(a+b)\).

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

    Definition \(\PageIndex{28}\)

    A complex conjugate pair is of the form \(a+bi,a+bi\).

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

    Example \(\PageIndex{29}\)

    Multiply: \((3-2 i)(3+2 i)\).

    Solution:

    Use FOIL

    Combine like terms and simplify \(i^{2}\).

    \(9-4(-1)\)

    Multiply and combine like terms.

    \(13\)

    Try It \(\PageIndex{30}\)

    Multiply: \((4-3 i) \cdot(4+3 i)\).

    Answer

    \(25\)

    Try It \(\PageIndex{31}\)

    Multiply: \((-2+5 i) \cdot(-2-5 i)\).

    Answer

    \(29\)

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

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

    The quantity a minus b in parentheses times the quantity a plus b in parentheses is written above the expression showing the product of 3 minus 2 i in parentheses and 3 plus 2 i in parentheses. In the next line a squared minus b squared is written above the expression 3 squared minus the quantity 2 i in parentheses squared. Simplifying we get 9 minus 4 i squared. This is equal to 9 minus 4 times negative 1. The final result is 13.
     

    Notice this is the same result we found in Example 8.8.9.

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

    \(\begin{array}{c}{(a-b i)(a+b i)} \\ {a^{2}-(b i)^{2}} \\ {a^{2}-b^{2} i^{2}} \\ {a^{2}-b^{2}(-1)} \\ {a^{2}+b^{2}}\end{array}\)

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

    Product of Complex Conjugates

    If \(a\) and \(b\) are real numbers, then

    \((a-b i)(a+b i)=a^{2}+b^{2}.\)

    Example \(\PageIndex{32}\)

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

    Solution:

      .
    Use the Product of Complex Conjugates Pattern, \((a-b i)(a+b i)=a^{2}+b^{2}\). .
    Simplify the squares. .
    Add. .
    Table 8.8.3
    Try It \(\PageIndex{33}\)

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

    Answer

    \(109\)

    Try It \(\PageIndex{34}\)

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

    Answer

    \(41\)

    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 \(\PageIndex{35}\)

    Divide: \(\dfrac{4+3 i}{3-4 i}\).

    Solution:

    Step 1: Write both the numerator and denominator in standard form. They are both in standard form.  
    Step 2: Multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of \(3-4i\) is \(3+4i\).  
    Step 3: Simplify and write the result in standard form.

    Use the pattern \((a-b i)(a+b i)=a^{2}+b^{2}\) in the denominator.

    Combine like terms.

    Simplify.

    Write the result in standard form.

    \(\begin{array}{c}{\dfrac{12+16 i+9 i+12 i^{2}}{9+16}} \\ {\dfrac{12+25 i-12}{25}} \\ {\dfrac{25 i}{25}} \\ {i}\end{array}\)
    Table 8.8.4
    Try It \(\PageIndex{36}\)

    Divide: \(\dfrac{2+5 i}{5-2 i}\).

    Answer

    \(i\)

    Try It \(\PageIndex{37}\)

    Divide: \(\dfrac{1+6 i}{6-i}\).

    Answer

    \(i\)

    We summarize the steps here.

    How to Divide Complex Numbers

    1. Write both the numerator and denominator in standard form.
    2. Multiply both the numerator and denominator by the complex conjugate of the denominator.
    3. Simplify and write the result in standard form.
    Example \(\PageIndex{38}\)

    Divide, writing the answers in standard form: \(\dfrac{-3}{5+2 i}\).

    Solution:

    \(\dfrac{-3}{5+2 i}\)

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

    Multiply in the numerator and use the Product of Complex Conjugates Pattern in the denominator.

    \(\dfrac{-15+6 i}{5^{2}+2^{2}}\)

    Simplify.

    \(\dfrac{-15+6 i}{29}\)

    Write in standard form.

    \(-\dfrac{15}{29}+\dfrac{6}{29} i\)

    Try It \(\PageIndex{39}\)

    Divide, writing the answer in standard form: \(\dfrac{4}{1-4 i}\).

    Answer

    \(\dfrac{4}{17}+\dfrac{16}{17} i\)

    Try It \(\PageIndex{40}\)

    Divide, writing the answer in standard form: \(\dfrac{-2}{-1+2 i}\).

    Answer

    \(\dfrac{2}{5}+\dfrac{4}{5} i\)

    Be careful as you find the conjugate of the denominator.

    Example \(\PageIndex{41}\)

    Divide: \(\dfrac{5+3 i}{4 i}\).

    Solution:

    \(\dfrac{5+3 i}{4 i}\)

    Write the denominator in standard form.

    \(\dfrac{5+3 i}{0+4 i}\)

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

    Simplify.

    Multiply.

    Simplify the \(i^{2}\).

    Rewrite in standard form.

    \(\dfrac{12}{16}-\dfrac{20}{16} i\)

    Simplify the fractions.

    \(\dfrac{3}{4}-\dfrac{5}{4} i\)

    Try It \(\PageIndex{42}\)

    Divide: \(\dfrac{3+3 i}{2 i}\).

    Answer

    \(\dfrac{3}{2}-\dfrac{3}{2} i\)

    Try It \(\PageIndex{43}\)

    Divide: \(\dfrac{2+4 i}{5 i}\).

    Answer

    \(\dfrac{4}{5}-\dfrac{2}{5} i\)

    Simplify Powers of \(i\)

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

    \(\begin{array}{ccc}{i^{1}} & {i^{2}} & {i^{3}} & {i^{4}} \\ {i} & {-1} & {i^{2}\cdot i} & {i^{2}\cdot i^{2}}\\ {}&{}&{-1\cdot i}&{(-1)(-1)}\\ {}&{}&{-i}&{1}\end{array}\)

    \(\begin{array}{cccc}{i^{5}} & {i^{6}} & {i^{7}} & {i^{8}} \\ {i^{4} \cdot i} & {i^{4} \cdot i^{2}} & {i^{4} \cdot i^{3}} & {i^{4} \cdot i^{4}} \\ {1 \cdot i} & {1 \cdot i^{2}} & {1 \cdot i^{3}} & {1 \cdot 1} \\ {i} & {i^{2}} & {i^{3}} & {1} \\ {}&{-1} & {-i}\end{array}\)

    We summarize this now.

    \(\begin{array}{ll}{i^{1}=i} & {i^{5}=i} \\ {i^{2}=-1} & {i^{6}=-1} \\ {i^{3}=-i} & {i^{7}=-i} \\ {i^{4}=1} & {i^{8}=1}\end{array}\)

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

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

    .
    Figure 8.8.13
    Example \(\PageIndex{44}\)

    Simplify: \(i^{86}\).

    Solution:

    \(i^{86}\)

    Divide \(86\) by \(4\) and rewrite \(i^{86}\) in the \(i^{n}=\left(i^{4}\right)^{q} \cdot i^{r}\) form.

    \(\left(1^{4}\right)^{21} \cdot i^{2}\)

    .
    Figure 8.8.14

    Simplify.

    \((1)^{21} \cdot(-1)\)

    Simplify.

    \(-1\)

    Try It \(\PageIndex{45}\)

    Simplify: \(i^{75}\).

    Answer

    \(-1\)

    Try It \(\PageIndex{46}\)

    Simplify: \(i^{92}\).

    Answer

    \(1\)

    Key Concepts

    • Square Root of a Negative Number
      • If \(b\) is a positive real number, then \(\sqrt{-b}=\sqrt{b} i\
      \(a+bi\)  
    \(b=0\)

    \(a+0\cdot i\)

    \(a\)

    Real number
    \(b\neq 0\) \(a+bi\) Imaginary number
    \(a=0\)

    \(0+bi\)

    \(bi\)

    Pure imaginary number
    Table 8.8.1
      • A complex number is in standard form when written as a + bi, where a, b are real numbers.
        The diagram has a rectangle with the labels “Complex Numbers” and a plus b i. A second rectangle has the labels “Real Numbers”, a plus b i, b = 0. A third rectangle has the labels “Imaginary Numbers”, a plus b i, b not equal to 0. Arrows go from the Real Numbers rectangle and Imaginary Numbers rectangle and point toward the Complex Numbers rectangle.
        Figure 8.8.2
    • Product of Complex Conjugates
      • If \(a, b\) are real numbers, then
        \((a−bi)(a+bi)=a^{2}+b^{2}\)
    • How to Divide Complex Numbers
      1. Write both the numerator and denominator in standard form.
      2. Multiply the numerator and denominator by the complex conjugate of the denominator.
      3. Simplify and write the result in standard form.

    Glossary

    complex conjugate pair
    A complex conjugate pair is of the form \(a+bi, a-bi\).
    complex number
    A complex number is of the form \(a+bi\), where \(a\) and \(b\) are real numbers. We call \(a\) the real part and \(b\) the imaginary part.
    complex number system
    The complex number system is made up of both the real numbers and the imaginary numbers.
    imaginary unit
    The imaginary unit \(i\) is the number whose square is \(–1\). \(i^{2}=-1\) or \(i=\sqrt{−1}\).
    standard form
    A complex number is in standard form when written as \(a+bi\), where \(a, b\) are real numbers.

    Practice Makes Perfect

    Evaluate a Square Root of a Negative Number

    In the following exercises, write each expression in terms of \(i\) and simplify if possible.

    1. a. \(\sqrt{-16}\) b. \(\sqrt{-11}\) c. \(\sqrt{-8}\)
    2. a. \(\sqrt{-121}\) b. \(\sqrt{-1}\) c. \(\sqrt{-20}\)
    3. a. \(\sqrt{-100}\) b. \(\sqrt{-13}\) c. \(\sqrt{-45}\)
    4. a. \(\sqrt{-49}\) b. \(\sqrt{-15}\) c. \(\sqrt{-75}\)
    Answer

    1. a. \(4i\)     b. \(i\sqrt{11}\)     c. \(2i\sqrt{2}\)

    3. a. \(10i\)     b. \(i\sqrt{13}\)     c. \(3i\sqrt{5}\)

    Add or Subtract Complex Numbers

    In the following exercises, add or subtract, putting the answer in \(a + bi\) form.

    5. \(\sqrt{-75}+\sqrt{-48}\)

    6. \(\sqrt{-12}+\sqrt{-75}\)

    7. \(\sqrt{-50}+\sqrt{-18}\)

    8. \(\sqrt{-72}+\sqrt{-8}\)

    9. \((1+3 i)+(7+4 i)\)

    10. \((6+2 i)+(3-4 i)\)

    11. \((8-i)+(6+3 i)\)

    12. \((7-4 i)+(-2-6 i)\)

    13. \((1-4 i)-(3-6 i)\)

    14. \((8-4 i)-(3+7 i)\)

    15. \((6+i)-(-2-4 i)\)

    16. \((-2+5 i)-(-5+6 i)\)

    17. \((5-\sqrt{-36})+(2-\sqrt{-49})\)

    18. \((-3+\sqrt{-64})+(5-\sqrt{-16})\)

    19. \((-7-\sqrt{-50})-(-32-\sqrt{-18})\)

    20. \((-5+\sqrt{-27})-(-4-\sqrt{-48})\)

    Answer

    5. \(0+\left(9\sqrt{3}\right)i\)

    7. \(0+\left(8\sqrt{2}\right)i\)

    9. \(8+7i\)

    11. \(14+2i\)

    13. \(-2+2i\)

    15. \(8+5i\)

    17. \(7-13i\)

    19. \(25-\left(2 \sqrt{2}\right) i\)

    Multiply Complex Numbers

    In the following exercises, multiply, putting the answer in \(a+bi\) form.

    21. \(4 i(5-3 i)\)

    22. \(2 i(-3+4 i)\)

    23. \(-6 i(-3-2 i)\)

    24. \(-i(6+5 i)\)

    25. \((4+3 i)(-5+6 i)\)

    26. \((-2-5 i)(-4+3 i)\)

    27. \((-3+3 i)(-2-7 i)\)

    28. \((-6-2 i)(-3-5 i)\)

    Answer

    21. \(12+20i\)

    23. \(-12+18i\)

    25. \(-38+9 i\)

    27. \(27+15i\)

    Multiply Complex Numbers

    In the following exercises, multiply using the Product of Binomial Squares Pattern, putting the answer in \(a+bi\) form.

    29. \((3+4 i)^{2}\)

    30. \((-1+5 i)^{2}\)

    31. \((-2-3 i)^{2}\)

    32. \((-6-5 i)^{2}\)

    Answer

    29. \(-7+24i\)

    31. \(-5-12i\)

    Multiply Complex Numbers

    In the following exercises, multiply, putting the answer in \(a+bi\) form.

    33. \(\sqrt{-25} \cdot \sqrt{-36}\)

    34. \(\sqrt{-4} \cdot \sqrt{-16}\)

    35. \(\sqrt{-9} \cdot \sqrt{-100}\)

    36. \(\sqrt{-64} \cdot \sqrt{-9}\)

    37. \((-2-\sqrt{-27})(4-\sqrt{-48})\)

    38. \((5-\sqrt{-12})(-3+\sqrt{-75})\)

    39. \((2+\sqrt{-8})(-4+\sqrt{-18})\)

    40. \((5+\sqrt{-18})(-2-\sqrt{-50})\)

    41. \((2-i)(2+i)\)

    42. \((4-5 i)(4+5 i)\)

    43. \((7-2 i)(7+2 i)\)

    44. \((-3-8 i)(-3+8 i)\)

    Answer

    33. \(30i = 0 + 30i\)

    35. \(-30 = -30 + 0i\)

    37. \(-44+\left(4 \sqrt{3}\right) i\)

    39. \(-20-\left(2 \sqrt{2}\right) i\)

    41. \(5 = 5 + 0i\)

    43. \(53 = 53 + 0i\)

    Multiply Complex Numbers

    In the following exercises, multiply using the Product of Complex Conjugates Pattern.

    45. \((7-i)(7+i)\)

    46. \((6-5 i)(6+5 i)\)

    47. \((9-2 i)(9+2 i)\)

    48. \((-3-4 i)(-3+4 i)\)

    Answer

    45. \(50\)

    47. \(85\)

    Divide Complex Numbers

    In the following exercises, divide, putting the answer in \(a+bi\) form.

    49. \(\dfrac{3+4 i}{4-3 i}\)

    50. \(\dfrac{5-2 i}{2+5 i}\)

    51. \(\dfrac{2+i}{3-4 i}\)

    52. \(\dfrac{3-2 i}{6+i}\)

    53. \(\dfrac{3}{2-3 i}\)

    54. \(\dfrac{2}{4-5 i}\)

    55. \(\dfrac{-4}{3-2 i}\)

    56. \(\dfrac{-1}{3+2 i}\)

    57. \(\dfrac{1+4 i}{3 i}\)

    58. \(\dfrac{4+3 i}{7 i}\)

    59. \(\dfrac{-2-3 i}{4 i}\)

    60. \(\dfrac{-3-5 i}{2 i}\)

    Answer

    49. \(i = 0 + i\)

    51. \(\dfrac{2}{25}+\dfrac{11}{25} i\)

    53. \(\dfrac{6}{13}+\dfrac{9}{13} i\)

    55. \(-\dfrac{12}{13}-\dfrac{8}{13} i\)

    57. \(\dfrac{4}{3}-\dfrac{1}{3} i\)

    59. \(-\dfrac{3}{4}+\dfrac{1}{2} i\)

    Simplify Powers of \(i\)

    In the following exercises, simplify.

    61. \(i^{41}\)

    62. \(i^{39}\)

    63. \(i^{66}\)

    64. \(i^{48}\)

    65. \(i^{128}\)

    66. \(i^{162}\)

    67. \(i^{137}\)

    68. \(i^{255}\)

    Answer

    61. \(i^{41} = i^{40}\cdot i = \left(i^{4}\right)^{10}\cdot i= i\)

    63. \(i^{66} = i^{64}\cdot i^{2} = \left(i^{4}\right)^{16}\cdot (-1)= -1\)

    65. \(i^{128} = \left(i^{4}\right)^{32} = 1\)

    67. \(i^{137} = i^{136}\cdot i = \left(i^{4}\right)^{34}\cdot i = 1 \cdot i = i\)

    Writing Exercises

    69. Explain the relationship between real numbers and complex numbers.

    70. Aniket multiplied as follows and he got the wrong answer. What is wrong with his reasoning?
    \(\begin{array}{c}{\sqrt{-7} \cdot \sqrt{-7}} \\ {\sqrt{49}} \\ {7}\end{array}\)

    71. Why is \(\sqrt{-64}=8 i\) but \(\sqrt[3]{-64}=-4\)?

    72. Explain how dividing complex numbers is similar to rationalizing a denominator.

    Answer

    69. Answers may vary

    71. Answers may vary

    Self Check

    a. After completing the Try It s, use this checklist to evaluate your mastery of the objectives of this section.

    The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “evaluate the square root of a negative number”, “add or subtract complex numbers”, “multiply complex numbers”, “divide complex numbers”, and “simplify powers of i”. The other columns are left blank so the learner can indicate their level of understanding.
     

    b. On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?


    6.7: Complex Numbers is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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