Complex Numbers and Their Operations

5.7 Complex Numbers and Their Operations

Learning Objectives

Define the imaginary unit and complex numbers.
Add and subtract complex numbers.
Multiply and divide complex numbers.

Introduction to Complex Numbers

Up to this point the square root of a negative number has been left undefined. For example, we know that $\sqrt{− 9}$ is not a real number.

$\sqrt{− 9} = ? or {(?)}^{2} = - 9$

There is no real number that when squared results in a negative number. We begin to resolve this issue by defining the imaginary unitDefined as $i = \sqrt{− 1}$ where $i^{2} = − 1 .$ , i, as the square root of −1.

$i = \sqrt{− 1} and i^{2} = − 1$

To express a square root of a negative number in terms of the imaginary unit i, we use the following property where a represents any non-negative real number:

$\sqrt{- a} = \sqrt{− 1 \cdot a} = \sqrt{− 1} \cdot \sqrt{a} = i \sqrt{a}$

With this we can write

$\sqrt{− 9} = \sqrt{− 1 \cdot 9} = \sqrt{− 1} \cdot \sqrt{9} = i \cdot 3 = 3 i$

If $\sqrt{− 9} = 3 i$ , then we would expect that $3 i$ squared will equal −9:

${(3 i)}^{2} = 9 i^{2} = 9 (− 1) = − 9 ✓$

In this way any square root of a negative real number can be written in terms of the imaginary unit. Such a number is often called an imaginary numberA square root of any negative real number..

Example 1

Rewrite in terms of the imaginary unit i.

$\sqrt{− 7}$
$\sqrt{− 25}$
$\sqrt{− 72}$

Solution:

$\sqrt{− 7} = \sqrt{− 1 \cdot 7} = \sqrt{− 1} \cdot \sqrt{7} = i \sqrt{7}$
$\begin{matrix} \sqrt{− 25} = \sqrt{− 1 \cdot 25} = \sqrt{− 1} \cdot \sqrt{25} = i \cdot 5 = 5 i \end{matrix}$
$\sqrt{− 72} = \sqrt{− 1 \cdot 36 \cdot 2} = \sqrt{− 1} \cdot \sqrt{36} \cdot \sqrt{2} = i \cdot 6 \cdot \sqrt{2} = 6 i \sqrt{2}$

Notation Note: When an imaginary number involves a radical, we place i in front of the radical. Consider the following:

$6 i \sqrt{2} = 6 \sqrt{2} i$

Since multiplication is commutative, these numbers are equivalent. However, in the form $6 \sqrt{2} i$ , the imaginary unit i is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place i in front of the radical and use $6 i \sqrt{2} .$

A complex numberA number of the form $a + b i,$ where a and b are real numbers. is any number of the form, $a + b i$ where a and b are real numbers. Here, a is called the real partThe real number a of a complex number $a + b i .$ and b is called the imaginary partThe real number b of a complex number $a + b i .$ . For example, $3 - 4 i$ is a complex number with a real part of 3 and an imaginary part of −4. It is important to note that any real number is also a complex number. For example, 5 is a real number; it can be written as $5 + 0 i$ with a real part of 5 and an imaginary part of 0. Hence, the set of real numbers, denoted $ℝ$ , is a subset of the set of complex numbers, denoted $ℂ .$

$ℂ = {a + b i | a, b \in ℝ}$

Complex numbers are used in many fields including electronics, engineering, physics, and mathematics. In this textbook we will use them to better understand solutions to equations such as $x^{2} + 4 = 0 .$ For this reason, we next explore algebraic operations with them.

Adding and Subtracting Complex Numbers

Adding or subtracting complex numbers is similar to adding and subtracting polynomials with like terms. We add or subtract the real parts and then the imaginary parts.

Example 2

Add: $(5 - 2 i) + (7 + 3 i) .$

Solution:

Add the real parts and then add the imaginary parts.

$\begin{matrix} (5 - 2 i) + (7 + 3 i) & = & 5 - 2 i + 7 + 3 i \\ = & 5 + 7 - 2 i + 3 i \\ = & 12 + i \end{matrix}$

Answer: $12 + i$

To subtract complex numbers, we subtract the real parts and subtract the imaginary parts. This is consistent with the use of the distributive property.

Example 3

Subtract: $(10 - 7 i) - (9 + 5 i) .$

Solution:

Distribute the negative sign and then combine like terms.

$\begin{matrix} (10 - 7 i) - (9 + 5 i) & = & 10 - 7 i - 9 - 5 i \\ = & 10 - 9 - 7 i - 5 i \\ = & 1 - 12 i \end{matrix}$

Answer: $1 - 12 i$

In general, given real numbers a, b, c and d:

$\begin{matrix} (a + b i) + (c + d i) & = & (a + c) + (b + d) i \\ (a + b i) - (c + d i) & = & (a - c) + (b - d) i \end{matrix}$

Example 4

Simplify: $(5 + i) + (2 - 3 i) - (4 - 7 i) .$

Solution:

$\begin{matrix} (5 + i) + (2 - 3 i) - (4 - 7 i) & = & 5 + i + 2 - 3 i - 4 + 7 i \\ = & 3 + 5 i \end{matrix}$

Answer: $3 + 5 i$

In summary, adding and subtracting complex numbers results in a complex number.

Multiplying and Dividing Complex Numbers

Multiplying complex numbers is similar to multiplying polynomials. The distributive property applies. In addition, we make use of the fact that $i^{2} = − 1$ to simplify the result into standard form $a + b i .$

Example 5

Multiply: $− 6 i (2 - 3 i) .$

Solution:

We begin by applying the distributive property.

$\begin{array}{l} − 6 i (2 - 3 i) & = & (− 6 i) \cdot 2 - (− 6 i) \cdot 3 i & D i s t r i b u t e . \\ = & − 12 i + 18 i^{2} & S u b s t i t u t e i^{2} = − 1 . \\ = & − 12 i + 18 (− 1) & S i m p l i f y . \\ = & − 12 i - 18 \\ = & − 18 - 12 i \end{array}$

Answer: $− 18 - 12 i$

Example 6

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

Solution:

$\begin{array}{l} (3 - 4 i) (4 + 5 i) & = & 3 \cdot 4 + 3 \cdot 5 i - 4 i \cdot 4 - 4 i \cdot 5 i & D i s t r i b u t e . \\ = & 12 + 15 i - 16 i - 20 i^{2} & S u b s t i t u t e i^{2} = − 1 . \\ = & 12 + 15 i - 16 i - 20 (− 1) \\ = & 12 - i + 20 \\ = & 32 - i \end{array}$

Answer: $32 - i$

In general, given real numbers a, b, c and d:

$\begin{matrix} (a + b i) (c + d i) & = & a c + a d i + b c i + b d i^{2} \\ = & a c + a d i + b c i + b d (− 1) \\ = & a c + (a d + b c) i - b d \\ = & (a c - b d) + (a d + b c) i \end{matrix}$

Try this! Simplify: ${(3 - 2 i)}^{2} .$

Answer: $5 - 12 i$

(click to see video)

Given a complex number $a + b i$ , its complex conjugateTwo complex numbers whose real parts are the same and imaginary parts are opposite. If given $a + b i$ , then its complex conjugate is $a - b i .$ is $a - b i .$ We next explore the product of complex conjugates.

Example 7

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

Solution:

$\begin{matrix} (5 + 2 i) (5 - 2 i) & = & 5 \cdot 5 - 5 \cdot 2 i + 2 i \cdot 5 - 2 i \cdot 2 i \\ = & 25 - 10 i + 10 i - 4 i^{2} \\ = & 25 - 4 (− 1) \\ = & 25 + 4 \\ = & 29 \end{matrix}$

Answer: 29

In general, the product of complex conjugatesThe real number that results from multiplying complex conjugates: $(a + b i) (a - b i) = a^{2} + b^{2} .$ follows:

$\begin{matrix} (a + b i) (a - b i) & = & a^{2} - a \cdot b i + b i \cdot a - b^{2} i^{2} \\ = & a^{2} - a b i + a b i - b^{2} (− 1) \\ = & a^{2} + b^{2} \end{matrix}$

Note that the result does not involve the imaginary unit; hence, it is real. This leads us to the very useful property

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

To divide complex numbers, we apply the technique used to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator. The result can then be simplified into standard form $a + b i .$

Example 8

Divide: $\frac{1}{2 - 3 i} .$

Solution:

In this example, the conjugate of the denominator is $2 + 3 i .$ Therefore, we will multiply by 1 in the form $\frac{(2 + 3 i)}{(2 + 3 i)} .$

$\begin{matrix} \frac{1}{2 - 3 i} & = & \frac{1}{(2 - 3 i)} \cdot \frac{(2 + 3 i)}{(2 + 3 i)} \\ = & \frac{(2 + 3 i)}{2^{2} + 3^{2}} \\ = & \frac{2 + 3 i}{4 + 9} \\ = & \frac{2 + 3 i}{13} \end{matrix}$

To write this complex number in standard form, we make use of the fact that 13 is a common denominator.

$\begin{matrix} \frac{2 + 3 i}{13} & = & \frac{2}{13} + \frac{3 i}{13} \\ = & \frac{2}{13} + \frac{3}{13} i \end{matrix}$

Answer: $\frac{2}{13} + \frac{3}{13} i$

Example 9

Divide: $\frac{1 - 5 i}{4 + i} .$

Solution:

$\begin{matrix} \frac{1 - 5 i}{4 + i} & = & \frac{(1 - 5 i)}{(4 + i)} \cdot \frac{(4 - i)}{(4 - i)} \\ = & \frac{4 - i - 20 i + 5 i^{2}}{4^{2} + 1^{2}} \\ = & \frac{4 - 21 i + 5 (− 1)}{16 + 1} \\ = & \frac{4 - 21 i - 5}{17} \\ = & \frac{− 1 - 21 i}{17} \\ = & - \frac{1}{17} - \frac{21}{17} i \end{matrix}$

Answer: $- \frac{1}{17} - \frac{21}{17} i$

In general, given real numbers a, b, c and d where c and d are not both 0:

$\begin{matrix} \frac{(a + b i)}{(c + d i)} & = & \frac{(a + b i)}{(c + d i)} \cdot \frac{(c - d i)}{(c - d i)} \\ = & \frac{a c - a d i + b c i - b d i^{2}}{c^{2} + d^{2}} \\ = & \frac{(a c + b d) + (b c - a d) i}{c^{2} + d^{2}} \\ = & (\frac{a c + b d}{c^{2} + d^{2}}) + (\frac{b c - a d}{c^{2} + d^{2}}) i \end{matrix}$

Example 10

Divide: $\frac{8 - 3 i}{2 i} .$

Solution:

Here we can think of $2 i = 0 + 2 i$ and thus we can see that its conjugate is $− 2 i = 0 - 2 i .$

$\begin{matrix} \frac{8 - 3 i}{2 i} & = & \frac{(8 - 3 i)}{(2 i)} \cdot \frac{(− 2 i)}{(− 2 i)} \\ = & \frac{− 16 i + 6 i^{2}}{− 4 i^{2}} \\ = & \frac{− 16 i + 6 (− 1)}{− 4 (− 1)} \\ = & \frac{− 16 i - 6}{4} \\ = & \frac{− 6 - 16 i}{4} \\ = & \frac{− 6}{4} - \frac{16 i}{4} \\ = & - \frac{3}{2} - 4 i \end{matrix}$

Because the denominator is a monomial, we could multiply numerator and denominator by 1 in the form of $\frac{i}{i}$ and save some steps reducing in the end.

$\begin{matrix} \frac{8 - 3 i}{2 i} & = & \frac{(8 - 3 i)}{(2 i)} \cdot \frac{i}{i} \\ = & \frac{8 i - 3 i^{2}}{2 i^{2}} \\ = & \frac{8 i - 3 (− 1)}{2 (− 1)} \\ = & \frac{8 i + 3}{− 2} \\ = & \frac{8 i}{− 2} + \frac{3}{− 2} \\ = & − 4 i - \frac{3}{2} \end{matrix}$

Answer: $- \frac{3}{2} - 4 i$

Try this! Divide: $\frac{3 + 2 i}{1 - i} .$

Answer: $\frac{1}{2} + \frac{5}{2} i$

(click to see video)

When multiplying and dividing complex numbers we must take care to understand that the product and quotient rules for radicals require that both a and b are positive. In other words, if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are both real numbers then we have the following rules.

$\begin{array}{l} Product rule for radicals : & \sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b} \\ Quotient rule for radicals : & \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{array}$

For example, we can demonstrate that the product rule is true when a and b are both positive as follows:

$\begin{array}{l} \sqrt{4} \cdot \sqrt{9} & = & \sqrt{36} \\ 2 \cdot 3 & = & 6 \\ 6 & = & 6 ✓ \end{array}$

However, when a and b are both negative the property is not true.

$\begin{array}{l} \sqrt{− 4} \cdot \sqrt{− 9} & \overset{?}{=} & \sqrt{36} \\ 2 i \cdot 3 i & = & 6 \\ 6 i^{2} & = & 6 \\ − 6 & = & 6 ✗ \end{array}$

Here $\sqrt{− 4}$ and $\sqrt{− 9}$ both are not real numbers and the product rule for radicals fails to produce a true statement. Therefore, to avoid some common errors associated with this technicality, ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Example 11

Multiply: $\sqrt{− 6} \cdot \sqrt{− 15} .$

Solution:

Begin by writing the radicals in terms of the imaginary unit i.

$\sqrt{− 6} \cdot \sqrt{− 15} = i \sqrt{6} \cdot i \sqrt{15}$

Now the radicands are both positive and the product rule for radicals applies.

$\begin{matrix} \sqrt{− 6} \cdot \sqrt{− 15} & = & i \sqrt{6} \cdot i \sqrt{15} \\ = & i^{2} \sqrt{6 \cdot 15} \\ = & (− 1) \sqrt{90} \\ = & (− 1) \sqrt{9 \cdot 10} \\ = & (− 1) \cdot 3 \cdot \sqrt{10} \\ = & - 3 \sqrt{10} \end{matrix}$

Answer: $− 3 \sqrt{10}$

Example 12

Multiply: $\sqrt{− 10} (\sqrt{− 6} - \sqrt{10}) .$

Solution:

Begin by writing the radicals in terms of the imaginary unit $i$ and then distribute.

$\begin{matrix} \sqrt{− 10} (\sqrt{− 6} - \sqrt{10}) & = & i \sqrt{10} (i \sqrt{6} - \sqrt{10}) \\ = & i^{2} \sqrt{60} - i \sqrt{100} \\ = & (− 1) \sqrt{4 \cdot 15} - i \sqrt{100} \\ = & (− 1) \cdot 2 \cdot \sqrt{15} - i \cdot 10 \\ = & - 2 \sqrt{15} - 10 i \end{matrix}$

Answer: $− 2 \sqrt{15} - 10 i$

In summary, multiplying and dividing complex numbers results in a complex number.

Try this! Simplify: ${(2 i \sqrt{2})}^{2} - {(3 - i \sqrt{5})}^{2} .$

Answer: $− 12 + 6 i \sqrt{5}$

(click to see video)

Key Takeaways

The imaginary unit i is defined to be the square root of negative one. In other words, $i = \sqrt{− 1}$ and $i^{2} = − 1 .$
Complex numbers have the form $a + b i$ where a and b are real numbers.
The set of real numbers is a subset of the complex numbers.
The result of adding, subtracting, multiplying, and dividing complex numbers is a complex number.
The product of complex conjugates, $a + b i$ and $a - b i$ , is a real number. Use this fact to divide complex numbers. Multiply the numerator and denominator of a fraction by the complex conjugate of the denominator and then simplify.
Ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Topic Exercises

Part A: Introduction to Complex Numbers

Rewrite in terms of imaginary unit i.

$\sqrt{− 81}$
$\sqrt{− 64}$
$- \sqrt{− 4}$
$- \sqrt{− 36}$
$\sqrt{− 20}$
$\sqrt{− 18}$
$\sqrt{− 50}$
$\sqrt{− 48}$
$- \sqrt{− 45}$
$- \sqrt{− 8}$
$\sqrt{- \frac{1}{16}}$
$\sqrt{- \frac{2}{9}}$
$\sqrt{− 0.25}$
$\sqrt{− 1.44}$

Write the complex number in standard form $a + b i .$

$5 - 2 \sqrt{− 4}$
$3 - 5 \sqrt{− 9}$
$− 2 + 3 \sqrt{− 8}$
$4 - 2 \sqrt{− 18}$
$\frac{3 - \sqrt{− 24}}{6}$
$\frac{2 + \sqrt{− 75}}{10}$
$\frac{\sqrt{− 63} - \sqrt{5}}{− 12}$
$\frac{- \sqrt{− 72} + \sqrt{8}}{− 24}$

Given that $i^{2} = − 1$ compute the following powers of $i .$

$i^{3}$
$i^{4}$
$i^{5}$
$i^{6}$
$i^{15}$
$i^{24}$

Part B: Adding and Subtracting Complex Numbers

Perform the operations.

$(3 + 5 i) + (7 - 4 i)$
$(6 - 7 i) + (− 5 - 2 i)$
$(− 8 - 3 i) + (5 + 2 i)$
$(− 10 + 15 i) + (15 - 20 i)$
$(\frac{1}{2} + \frac{3}{4} i) + (\frac{1}{6} - \frac{1}{8} i)$
$(\frac{2}{5} - \frac{1}{6} i) + (\frac{1}{10} - \frac{3}{2} i)$
$(5 + 2 i) - (8 - 3 i)$
$(7 - i) - (− 6 - 9 i)$
$(− 9 - 5 i) - (8 + 12 i)$
$(− 11 + 2 i) - (13 - 7 i)$
$(\frac{1}{14} + \frac{3}{2} i) - (\frac{4}{7} - \frac{3}{4} i)$
$(\frac{3}{8} - \frac{1}{3} i) - (\frac{1}{2} - \frac{1}{2} i)$
$(2 - i) + (3 + 4 i) - (6 - 5 i)$
$(7 + 2 i) - (6 - i) - (3 - 4 i)$
$(\frac{1}{3} - i) - (1 - \frac{1}{2} i) - (\frac{1}{6} + \frac{1}{6} i)$
$(1 - \frac{3}{4} i) + (\frac{5}{2} + i) - (\frac{1}{4} - \frac{5}{8} i)$
$(5 - 3 i) - (2 + 7 i) - (1 - 10 i)$
$(6 - 11 i) + (2 + 3 i) - (8 - 4 i)$
$\sqrt{− 16} - (3 - \sqrt{− 1})$
$\sqrt{− 100} + (\sqrt{− 9} + 7)$
$(1 + \sqrt{− 1}) - (1 - \sqrt{− 1})$
$(3 - \sqrt{− 81}) - (5 - 3 \sqrt{− 9})$
$(5 - 2 \sqrt{− 25}) - (− 3 + 4 \sqrt{− 1})$
$(− 12 - \sqrt{− 1}) - (3 - \sqrt{− 49})$

Part C: Multiplying and Dividing Complex Numbers

Perform the operations.

$i (1 - i)$
$i (1 + i)$
$2 i (7 - 4 i)$
$6 i (1 - 2 i)$
$− 2 i (3 - 4 i)$
$− 5 i (2 - i)$
$(2 + i) (2 - 3 i)$
$(3 - 5 i) (1 - 2 i)$
$(1 - i) (8 - 9 i)$
$(1 + 5 i) (5 + 2 i)$
${(4 + 3 i)}^{2}$
${(− 1 + 2 i)}^{2}$
${(2 - 5 i)}^{2}$
${(5 - i)}^{2}$
$(1 + i) (1 - i)$
$(2 - i) (2 + i)$
$(4 - 2 i) (4 + 2 i)$
$(6 + 5 i) (6 - 5 i)$
$(\frac{1}{2} + \frac{2}{3} i) (\frac{1}{3} - \frac{1}{2} i)$
$(\frac{2}{3} - \frac{1}{3} i) (\frac{1}{2} - \frac{3}{2} i)$
${(2 - i)}^{3}$
${(1 - 3 i)}^{3}$
$\sqrt{− 2} (\sqrt{− 2} - \sqrt{6})$
$\sqrt{− 1} (\sqrt{− 1} + \sqrt{8})$
$\sqrt{− 6} (\sqrt{10} - \sqrt{− 6})$
$\sqrt{− 15} (\sqrt{3} - \sqrt{− 10})$
$(2 - 3 \sqrt{− 2}) (2 + 3 \sqrt{− 2})$
$(1 + \sqrt{− 5}) (1 - \sqrt{− 5})$
$(1 - 3 \sqrt{− 4}) (2 + \sqrt{− 9})$
$(2 - 3 \sqrt{− 1}) (1 + 2 \sqrt{− 16})$
$(2 - 3 i \sqrt{2}) (3 + i \sqrt{2})$
$(− 1 + i \sqrt{3}) (2 - 2 i \sqrt{3})$
$\frac{− 3}{i}$
$\frac{5}{i}$
$\frac{1}{5 + 4 i}$
$\frac{1}{3 - 4 i}$
$\frac{15}{1 - 2 i}$
$\frac{29}{5 + 2 i}$
$\frac{20 i}{1 - 3 i}$
$\frac{10 i}{1 + 2 i}$
$\frac{10 - 5 i}{3 - i}$
$\frac{5 - 2 i}{1 - 2 i}$
$\frac{5 + 10 i}{3 + 4 i}$
$\frac{2 - 4 i}{5 + 3 i}$
$\frac{26 + 13 i}{2 - 3 i}$
$\frac{4 + 2 i}{1 + i}$
$\frac{3 - i}{2 i}$
$\frac{− 5 + 2 i}{4 i}$
$\frac{1}{a - b i}$
$\frac{i}{a + b i}$
$\frac{1 - \sqrt{− 1}}{1 + \sqrt{− 1}}$
$\frac{1 + \sqrt{− 9}}{1 - \sqrt{− 9}}$
$\frac{- \sqrt{− 6}}{\sqrt{18} + \sqrt{− 4}}$
$\frac{\sqrt{− 12}}{\sqrt{2} - \sqrt{− 27}}$

Given that $i^{- n} = \frac{1}{i^{n}}$ compute the following powers of $i .$

$i^{− 1}$
$i^{− 2}$
$i^{− 3}$
$i^{− 4}$

Perform the operations and simplify.

$2 i (2 - i) - i (3 - 4 i)$
$i (5 - i) - 3 i (1 - 6 i)$
$5 - 3 {(1 - i)}^{2}$
$2 {(1 - 2 i)}^{2} + 3 i$
${(1 - i)}^{2} - 2 (1 - i) + 2$
${(1 + i)}^{2} - 2 (1 + i) + 2$
${(2 i \sqrt{2})}^{2} + 5$
${(3 i \sqrt{5})}^{2} - {(i \sqrt{3})}^{2}$
${(\sqrt{2} - i)}^{2} - {(\sqrt{2} + i)}^{2}$
${(i \sqrt{3} + 1)}^{2} - {(4 i \sqrt{2})}^{2}$
${(\frac{1}{1 + i})}^{2}$
${(\frac{1}{1 + i})}^{3}$
${(a - b i)}^{2} - {(a + b i)}^{2}$
$(a^{2} + a i + 1) (a^{2} - a i + 1)$
Show that both $− 2 i$ and $2 i$ satisfy $x^{2} + 4 = 0 .$
Show that both $- i$ and $i$ satisfy $x^{2} + 1 = 0 .$
Show that both $3 - 2 i$ and $3 + 2 i$ satisfy $x^{2} - 6 x + 13 = 0 .$
Show that both $5 - i$ and $5 + i$ satisfy $x^{2} - 10 x + 26 = 0 .$
Show that 3, $- 2 i$ , and $2 i$ are all solutions to $x^{3} - 3 x^{2} + 4 x - 12 = 0 .$
Show that −2, $1 - i$ , and $1 + i$ are all solutions to $x^{3} - 2 x + 4 = 0 .$

Part D: Discussion Board.

Research and discuss the history of the imaginary unit and complex numbers.
How would you define $i^{0}$ and why?
Research what it means to calculate the absolute value of a complex number $| a + b i | .$ Illustrate your finding with an example.
Explore the powers of i. Look for a pattern and share your findings.

Answers

$9 i$
$− 2 i$
$2 i \sqrt{5}$
$5 i \sqrt{2}$
$− 3 i \sqrt{5}$
$\frac{i}{4}$
$0.5 i$
$5 - 4 i$
$− 2 + 6 i \sqrt{2}$
$\frac{1}{2} - \frac{\sqrt{6}}{3} i$
$\frac{\sqrt{5}}{12} - \frac{\sqrt{7}}{4} i$
$- i$
$i$
$- i$

$10 + i$
$− 3 - i$
$\frac{2}{3} + \frac{5}{8} i$
$− 3 + 5 i$
$− 17 - 17 i$
$- \frac{1}{2} + \frac{9}{4} i$
$− 1 + 8 i$
$- \frac{5}{6} - \frac{2}{3} i$
2
$− 3 + 5 i$
$2 i$
$8 - 14 i$

$1 + i$
$8 + 14 i$
$− 8 - 6 i$
$7 - 4 i$
$− 1 - 17 i$
$7 + 24 i$
$− 21 - 20 i$
2
20
$\frac{1}{2} - \frac{1}{36} i$
$2 - 11 i$
$− 2 - 2 i \sqrt{3}$
$6 + 2 i \sqrt{15}$
22
$20 - 9 i$
$12 - 7 i \sqrt{2}$
$3 i$
$\frac{5}{41} - \frac{4}{41} i$
$3 + 6 i$
$− 6 + 2 i$
$\frac{7}{2} - \frac{1}{2} i$
$\frac{11}{5} - \frac{2}{5} i$
$1 + 8 i$
$- \frac{1}{2} - \frac{3}{2} i$
$\frac{a}{a^{2} + b^{2}} + \frac{b}{a^{2} + b^{2}} i$
$- i$
$- \frac{\sqrt{6}}{11} - \frac{3 \sqrt{3}}{11} i$
$- i$
$i$
$− 2 + i$
$5 + 6 i$
0
−3
$− 4 i \sqrt{2}$
$- \frac{i}{2}$
$− 4 a b i$
Proof
Proof
Proof

Answer may vary
Answer may vary