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# Imaginary Numbers Explained with Formula: $$\sqrt{-1} \text{= ?}$$

## What they are and how to simplify

Part I. Simplifying Powers of $$i$$

$$i^5 = ? \\ i ^ {21} = ?$$

### Video Tutorial on Simplifying Imaginary Numbers

#### What is an imaginary number anyway?

Imaginary numbers are based on the mathematical number $$i$$.

$$i \text { is defined to be } \sqrt{-1}$$

From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples.

Table 1

$\text{ Table 1} \\ \begin{array}{ccc|c} \hline Expression & & Work & Result \\\hline \red{i^ \textbf{2}} & = & i \cdot i = \sqrt{-1} \cdot \sqrt{-1} & \red{ \textbf{ -1 }} \\\hline \red{i^ \textbf{3}} & = & i^2 \cdot i = -1 \cdot i & \red{ \textbf{-i} } \\\hline \red{ i^ \textbf{4} } & = & i^2 \cdot i^2 -1 \cdot -1 = & \red{1} \\\hline \end{array}$

You should understand Table 1 above .

Table 1 above boils down to the 4 conversions that you can see in Table 2 below. You should memorize Table 2 below because once you start actually solving problems, you'll see you use table 2 over and over again!

Table 2

$\text{ Table 2}$

#### What is the larger pattern?

In order to understand how to simplify the powers of $$i$$, let's look at some more examples, and we'll soon see a formula emerge!

$\begin{array}{c|c|c} Expression & Work & Result \\\hline \red{i^ \textbf{5}} & \blue{i^4} \cdot i^1 = \blue{1} \cdot i & \red{ \textbf{ i }} \\\hline \red{i^ \textbf{6}} & \blue{i^4} \cdot i^2= \blue{1} \cdot -1 & \red{ \textbf{-1}} \\\hline \red{ i^ \textbf{7} } & \blue{ i^4} \cdot i^3 =\blue{1} \cdot -i & \red{ \boldsymbol{ -i}} \\\hline \red{ i^ \textbf{8} } & = \blue{ i^4} \cdot \blue{ i^4}= \blue{1} \cdot \blue{1} = & \red{ \textbf{ 1}} \\\hline \end{array}$

Do you see the pattern yet? Let's look at 4 more and then summarize.

$\begin{array}{c|c|c} Expression & Work & Result \\\hline \red{i^ \textbf{9}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^1 = \blue{1} \cdot \blue{1} \cdot i = & \red{ \textbf{ i }} \\\hline \red{i^ \textbf{10}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^2 = \blue{1} \cdot \blue{1} \cdot i^2 = & \red{ \textbf{ -1 }} \\\hline \red{i^ \textbf{11}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^3 = \blue{1} \cdot \blue{1} \cdot i^3 = & \red{ \textbf{ -i }} \\\hline \red{i^ \textbf{12}} & = \blue{i^4} \cdot \blue{i^4} \cdot \blue{i^4} = \blue{1} \cdot \blue{1} \cdot \blue{1}= & \red{ \textbf{ 1 }} \\\hline \end{array}$

### Practice Problems

##### Problem 1
Step 1

Calculate the remainder

$$23 \div 4$$ has a remainder of $$\red{3}$$

Step 2

Rewrite

$$i^3$$

Step 3

Simplify using Table2

$$i^{\red{3}} = -i$$

##### Problem 2
Step 1

Calculate the remainder

$$18 \div 4$$ has a remainder of $$\red{2}$$

Step 2

Rewrite

$$i^{ \red{2}}$$

Step 3

Simplify using Table2

$$i^{ \red{2}} = -1$$

##### Problem 3
Step 1

Calculate the remainder

$$41 \div 4$$ has a remainder of $$\red{1}$$

Step 2

Rewrite

$$i^{ \red{1}}$$

Step 3

Simplify using Table2

$$i^{ \red{1}} = i$$

##### Problem 4
Step 1

Calculate the remainder

$$100 \div 4$$ has a remainder of $$\red{0}$$

Step 2

Rewrite

$$i^{ \red{0}}$$

Step 3

Simplify using Table2

$$i^{ \red{0}} = 1$$

##### Problem 5

Remember your order of operations. Exponents must be evaluated before multiplication so you can think of this problem as $$5 \cdot (\color{Blue}{i^ {22}})$$

Step 1

Calculate the remainder

$$22 \div 4$$ has a remainder of $$\red{2}$$

Step 2

Rewrite

$$i^{ \red{2}}$$

Step 3

Simplify using Table2

$$i^{ \red{2}} = -1$$

Step 4

$$5 \cdot ( {\color{Blue}-1} ) = -5$$

##### Problem 6

Remember your order of operations. Exponents must be evaluated before multiplication so you can think of this problem as $$12 \cdot ( {\color{Blue}i^ {36}})$$

Step 1

Calculate the remainder

$$36 \div 4$$ has a remainder of $$\red{0}$$

Step 2

Rewrite

$$i^{ \red{0}}$$

Step 3

Simplify using Table2

$$i^{ \red{0}} = 1$$

Step 4

$$12 \cdot ( {\color{Blue} 1} ) = 12$$

##### Problem 7

Remember your order of operations. Exponents must be evaluated before multiplication so you can think of this problem as $$7 \cdot ( {\color{Blue}i^ {103}})$$

Step 1

Calculate the remainder

$$103 \div 4$$ has a remainder of $$\red{3}$$

Step 2

Rewrite

$$i^{ \red{3}}$$

Step 3

Simplify using Table2

$$i^{ \red{3}} = -i$$

Step 4

$$7 \cdot ( {\color{Blue} -i} ) = -7i$$