Use the Complex Number System - Intermediate Algebra

8.8 Use the Complex Number System

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 8.22

Before you get started, take this readiness quiz.

Given the numbers $−4, - \sqrt{7}, 0. \bar{5}, \frac{7}{3}, 3, \sqrt{81},$ list the ⓐ rational numbers, ⓑ irrational numbers, ⓒ real numbers.
If you missed this problem, review Example 1.42.

Be Prepared 8.23

Multiply: $(x - 3) (2 x + 5) .$
If you missed this problem, review Example 5.28.

Be Prepared 8.24

Rationalize the denominator: $\frac{\sqrt{5}}{\sqrt{5} - \sqrt{3}} .$
If you missed this problem, review Example 5.32.

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 x² = –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.

Imaginary Unit

The imaginary unit i is the number whose square is –1.

i^{2} = −1 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

\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 8.76

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

ⓐ $\sqrt{−25}$ ⓑ $\sqrt{−7}$ ⓒ $\sqrt{−12} .$

Solution

ⓐ

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

Step-by-step simplification of the square root of -25.

ⓑ

	$\sqrt{−7}$
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.

Steps to simplify the square root of a negative number, showing the use of the imaginary unit 'i'.

ⓒ

	$\sqrt{−12}$
Use the definition of the square root of negative numbers.	$\sqrt{12} i$
Simplify $\sqrt{12} .$	$2 \sqrt{3} i$

Step-by-step simplification of sqrt(-12) into its complex number form, 2 sqrt(3)i.

Try It 8.151

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

ⓐ $\sqrt{−81}$ ⓑ $\sqrt{−5}$ ⓒ $\sqrt{−18} .$

Try It 8.152

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

ⓐ $\sqrt{−36}$ ⓑ $\sqrt{−3}$ ⓒ $\sqrt{−27} .$

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.

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 $a + b i,$ where a and b are real numbers.

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

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

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

We summarize this here.

	$a + b i$
$b = 0$	$\begin{matrix} a + 0 \cdot i \\ a \end{matrix}$	Real number
$b \neq 0$	$a + b i$	Imaginary number
$a = 0$	$\begin{array}{l} 0 + b i \\ b i \end{array}$	Pure imaginary number

The standard form of a complex number is $a + b i,$ 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”.

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 8.77

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.	$5 \sqrt{3} i$

Step-by-step solution for adding square roots of negative numbers, demonstrating simplification into complex numbers.

Try It 8.153

Add: $\sqrt{−8} + \sqrt{−32} .$

Try It 8.154

Add: $\sqrt{−27} + \sqrt{−48} .$

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

Example 8.78

Simplify: ⓐ $(4 - 3 i) + (5 + 6 i)$ ⓑ $(2 - 5 i) - (5 - 2 i) .$

Solution

ⓐ

	$(4 - 3 i) + (5 + 6 i)$
Use the Associative Property to put the real parts and the imaginary parts together.	$(4 + 5) + (−3 i + 6 i)$
Simplify.	$9 + 3 i$

Step-by-step example of adding complex numbers, illustrating the use of the associative property to simplify the expression.

ⓑ

	$(2 - 5 i) - (5 - 2 i)$
Distribute.	$2 - 5 i - 5 + 2 i$
Use the Associative Property to put the real parts and the imaginary parts together.	$2 - 5 - 5 i + 2 i$
Simplify.	$−3 - 3 i$

A step-by-step guide demonstrating the subtraction and simplification of two complex numbers.

Try It 8.155

Simplify: ⓐ $(2 + 7 i) + (4 - 2 i)$ ⓑ $(8 - 4 i) - (2 - i) .$

Try It 8.156

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

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 8.79

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

Solution

	$2 i (7 - 5 i)$
Distribute.	$14 i - 10 i^{2}$
Simplify $i^{2} .$	$14 i - 10 (−1)$
Multiply.	$14 i + 10$
Write in standard form.	$10 + 14 i$

Step-by-step simplification of the complex number expression 2i(7-5i) into standard form (10 + 14i).

Try It 8.157

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

Try It 8.158

Multiply: $−3 i (2 + 4 i) .$

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

Example 8.80

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

Solution

	$(3 + 2 i) (4 - 3 i)$
Use FOIL.	$12 - 9 i + 8 i - 6 i^{2}$
Simplify $i^{2}$ and combine like terms.	$12 - i - 6 (−1)$
Multiply.	$12 - i + 6$
Combine the real parts.	$18 - i$

A step-by-step guide to multiplying complex numbers using the FOIL method, detailing each calculation stage from the initial product to the simplified result.

Try It 8.159

Multiply: $(5 - 3 i) (−1 - 2 i) .$

Try It 8.160

Multiply: $(−4 - 3 i) (2 + i) .$

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

Example 8.81

Multiply: ${(3 + 2 i)}^{2}$

Solution

	$A mathematical expression showing the fraction (a + b) / (3 + 2i) squared. The numerator 'a + b' is displayed in red text.$
Use the Product of Binomial Squares Pattern, ${(a + b)}^{2} = a^{2} + 2 a b + b^{2} .$
Simplify.
Simplify $i^{2} .$
Simplify.

Try It 8.161

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

Try It 8.162

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

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 $\sqrt{− b} = \sqrt{b} i .$ This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Example 8.82

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.	$12 i^{2}$
Simplify $i^{2}$ and multiply.	$−12$

Step-by-step multiplication of square roots of negative numbers, demonstrating conversion to complex numbers and simplification.

Try It 8.163

Multiply: $\sqrt{−49} \cdot \sqrt{−4} .$

Try It 8.164

Multiply: $\sqrt{−36} \cdot \sqrt{−81} .$

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 8.83

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 $\sqrt{− b} = \sqrt{b} i .$	$(3 - 2 \sqrt{3} i) (5 + 3 \sqrt{3} i)$
Use FOIL.	$15 + 9 \sqrt{3} i - 10 \sqrt{3} i - 6 \cdot 3 i^{2}$
Combine like terms and simplify $i^{2} .$	$15 - \sqrt{3} i - 6 \cdot (−3)$
Multiply and combine like terms.	$33 - \sqrt{3} i$

Detailed steps demonstrating the multiplication of two complex numbers, simplifying to their final form.

Try It 8.165

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

Try It 8.166

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

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 $a + b i,$ its conjugate is $a - b i .$ 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 $a + b i,$ $a - b i .$

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

Example 8.84

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

Solution

	$(3 - 2 i) (3 + 2 i)$
Use FOIL.	$9 + 6 i - 6 i - 4 i^{2}$
Combine like terms and simplify $i^{2} .$	$9 - 4 (−1)$
Multiply and combine like terms.	13

Step-by-step multiplication and simplification of complex conjugates (3 - 2i)(3 + 2i) to a real number result.

Try It 8.167

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

Try It 8.168

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

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 8.84.

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

\begin{matrix} (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{matrix}

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 8.85

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.

Try It 8.169

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

Try It 8.170

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

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 8.86

How to Divide Complex Numbers

Divide: $\frac{4 + 3 i}{3 - 4 i} .$

Solution

Step 1 is to write both the numerator and denominator in standard form. For this example they are both in standard form.

Step 2 is to multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of 3 minus 4 i is 3 plus 4 i. The resulting expression is the quantity 4 plus 3 i in parentheses times the quantity 3 plus 4 i in parentheses divided by the product of 3 minus 4 i in parentheses and the quantity 3 plus 4 i in parentheses.

Step 3 is to simplify and write the result in standard form. Use the pattern the quantity a plus b i in parentheses equals a squared plus b squared in the denominator. The expression for this example then becomes the quantity 12 plus 16 i plus 9 i plus 12 i squared in parentheses divided by the sum of 9 and 16. Combining like terms we get the quantity 12 plus 25 i minus 12 in parentheses divided by 25. Simplifying we get 25 i divided by 25. Write the result in standard form. The result is i.

Try It 8.171

Divide: $\frac{2 + 5 i}{5 - 2 i} .$

Try It 8.172

Divide: $\frac{1 + 6 i}{6 - i} .$

We summarize the steps here.

How To

How to divide complex numbers.

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

Example 8.87

Divide, writing the answer in standard form: $\frac{−3}{5 + 2 i} .$

Solution

	$\frac{−3}{5 + 2 i}$
Multiply the numerator and denominator by the complex conjugate of the denominator.	$\frac{−3 (5 - 2 i)}{(5 + 2 i) (5 - 2 i)}$
Multiply in the numerator and use the Product of Complex Conjugates Pattern in the denominator.	$\frac{−15 + 6 i}{5^{2} + 2^{2}}$
Simplify.	$\frac{−15 + 6 i}{29}$
Write in standard form.	$- \frac{15}{29} + \frac{6}{29} i$

Step-by-step process for dividing complex numbers by rationalizing the denominator, showing actions and corresponding mathematical expressions.

Try It 8.173

Divide, writing the answer in standard form: $\frac{4}{1 - 4 i} .$

Try It 8.174

Divide, writing the answer in standard form: $\frac{−2}{−1 + 2 i} .$

Be careful as you find the conjugate of the denominator.

Example 8.88

Divide: $\frac{5 + 3 i}{4 i} .$

Solution

	$\frac{5 + 3 i}{4 i}$
Write the denominator in standard form.	$\frac{5 + 3 i}{0 + 4 i}$
Multiply the numerator and denominator by the complex conjugate of the denominator.	$\frac{(5 + 3 i) (0 - 4 i)}{(0 + 4 i) (0 - 4 i)}$
Simplify.	$\frac{(5 + 3 i) (−4 i)}{(4 i) (−4 i)}$
Multiply.	$\frac{−20 i - 12 i^{2}}{−16 i^{2}}$
Simplify the $i^{2} .$	$\frac{−20 i + 12}{16}$
Rewrite in standard form.	$\frac{12}{16} - \frac{20}{16} i$
Simplify the fractions.	$\frac{3}{4} - \frac{5}{4} i$

This table illustrates the step-by-step process for dividing complex numbers, transforming a complex fraction into its standard a + bi form.

Try It 8.175

Divide: $\frac{3 + 3 i}{2 i} .$

Try It 8.176

Divide: $\frac{2 + 4 i}{5 i} .$

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{matrix} 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 \\ 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.

\begin{matrix} 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{matrix}

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 i⁴ = 1, we rewrite each power, iⁿ, as a product using i⁴ to a power and another power of i.

We rewrite it in the form $i^{n} = {(i^{4})}^{q} \cdot i^{r},$ 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 i⁵⁷, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, $57 = 4 \cdot 14 + 1 .$ So we write $i^{57} = {(1^{4})}^{14} \cdot i^{1}$ and then simplify from there.

A mathematical problem demonstrating the simplification of i to the power of 57, using long division (57 / 4 = 14 remainder 1) to determine that i^57 equals i.

Example 8.89

Simplify: $i^{86} .$

Solution

	$i^{86}$
Divide 86 by 4 and rewrite $i^{86}$ in the $i^{n} = {(i^{4})}^{q} \cdot i^{r}$ form.	${(1^{4})}^{21} \cdot i^{2}$

Simplify.	${(1)}^{21} \cdot (−1)$
Simplify.	$–1$

This table illustrates the step-by-step simplification of the imaginary unit i raised to the power of 86.

Try It 8.177

Simplify: $i^{75} .$

Try It 8.178

Simplify: $i^{92} .$

Media

Access these online resources for additional instruction and practice with the complex number system.

Section 8.8 Exercises

Practice Makes Perfect

Evaluate the Square Root of a Negative Number

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

409.

ⓐ $\sqrt{−16}$ ⓑ $\sqrt{−11}$
ⓒ $\sqrt{−8}$

410.

ⓐ $\sqrt{−121}$ ⓑ $\sqrt{−1}$ ⓒ $\sqrt{−20}$

411.

ⓐ $\sqrt{−100}$ ⓑ $\sqrt{−13}$ ⓒ $\sqrt{−45}$

412.

ⓐ $\sqrt{−49}$ ⓑ $\sqrt{−15}$ ⓒ $\sqrt{−75}$

Add or Subtract Complex Numbers In the following exercises, add or subtract.

413.

$\sqrt{−75} + \sqrt{−48}$

414.

$\sqrt{−12} + \sqrt{−75}$

415.

$\sqrt{−50} + \sqrt{−18}$

416.

$\sqrt{−72} + \sqrt{−8}$

417.

$(1 + 3 i) + (7 + 4 i)$

418.

$(6 + 2 i) + (3 - 4 i)$

419.

$(8 - i) + (6 + 3 i)$

420.

$(7 - 4 i) + (−2 - 6 i)$

421.

$(1 - 4 i) - (3 - 6 i)$

422.

$(8 - 4 i) - (3 + 7 i)$

423.

$(6 + i) - (−2 - 4 i)$

424.

$(−2 + 5 i) - (−5 + 6 i)$

425.

$(5 - \sqrt{−36}) + (2 - \sqrt{−49})$

426.

$(−3 + \sqrt{−64}) + (5 - \sqrt{−16})$

427.

$(−7 - \sqrt{−50}) - (−32 - \sqrt{−18})$

428.

$(−5 + \sqrt{−27}) - (−4 - \sqrt{−48})$

Multiply Complex Numbers

In the following exercises, multiply.

429.

$4 i (5 - 3 i)$

430.

$2 i (−3 + 4 i)$

431.

$−6 i (−3 - 2 i)$

432.

$− i (6 + 5 i)$

433.

$(4 + 3 i) (−5 + 6 i)$

434.

$(−2 - 5 i) (−4 + 3 i)$

435.

$(−3 + 3 i) (−2 - 7 i)$

436.

$(−6 - 2 i) (−3 - 5 i)$

In the following exercises, multiply using the Product of Binomial Squares Pattern.

437.

${(3 + 4 i)}^{2}$

438.

${(−1 + 5 i)}^{2}$

439.

${(−2 - 3 i)}^{2}$

440.

${(−6 - 5 i)}^{2}$

In the following exercises, multiply.

441.

$\sqrt{−25} \cdot \sqrt{−36}$

442.

$\sqrt{−4} \cdot \sqrt{−16}$

443.

$\sqrt{−9} \cdot \sqrt{−100}$

444.

$\sqrt{−64} \cdot \sqrt{−9}$

445.

$(−2 - \sqrt{−27}) (4 - \sqrt{−48})$

446.

$(5 - \sqrt{−12}) (−3 + \sqrt{−75})$

447.

$(2 + \sqrt{−8}) (−4 + \sqrt{−18})$

448.

$(5 + \sqrt{−18}) (−2 - \sqrt{−50})$

449.

$(2 - i) (2 + i)$

450.

$(4 - 5 i) (4 + 5 i)$

451.

$(7 - 2 i) (7 + 2 i)$

452.

$(−3 - 8 i) (−3 + 8 i)$

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

453.

$(7 - i) (7 + i)$

454.

$(6 - 5 i) (6 + 5 i)$

455.

$(9 - 2 i) (9 + 2 i)$

456.

$(−3 - 4 i) (−3 + 4 i)$

Divide Complex Numbers

In the following exercises, divide.

457.

$\frac{3 + 4 i}{4 - 3 i}$

458.

$\frac{5 - 2 i}{2 + 5 i}$

459.

$\frac{2 + i}{3 - 4 i}$

460.

$\frac{3 - 2 i}{6 + i}$

461.

$\frac{3}{2 - 3 i}$

462.

$\frac{2}{4 - 5 i}$

463.

$\frac{−4}{3 - 2 i}$

464.

$\frac{−1}{3 + 2 i}$

465.

$\frac{1 + 4 i}{3 i}$

466.

$\frac{4 + 3 i}{7 i}$

467.

$\frac{−2 - 3 i}{4 i}$

468.

$\frac{−3 - 5 i}{2 i}$

Simplify Powers of i

In the following exercises, simplify.

469.

$i^{41}$

470.

$i^{39}$

471.

$i^{66}$

472.

$i^{48}$

473.

$i^{128}$

474.

$i^{162}$

475.

$i^{137}$

476.

$i^{255}$

Writing Exercises

477.

Explain the relationship between real numbers and complex numbers.

478.

Aniket multiplied as follows and he got the wrong answer. What is wrong with his reasoning?

$\begin{matrix} \sqrt{−7} \cdot \sqrt{−7} \\ \sqrt{49} \\ 7 \end{matrix}$

479.

Why is $\sqrt{−64} = 8 i$ but $\sqrt[3]{−64} = −4 .$

480.

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

Self Check

ⓐ After completing the exercises, 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.

ⓑ 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?