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

What they are and how to simplify

Part I. (this Page)

Simplifying Powers of $$i$$

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

Part II. Simplifying negative radicands

$$ \sqrt{-25} = ? \\ \sqrt{-18} = ? $$

What is an imaginary number anyway?

Imaginary numbers are based around 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
  • $$ \color{Red}{i^ \textbf{2}} = \sqrt{-1} \cdot \sqrt{-1} = $$
    $$ \color{Red}{ \textbf{ -1 }} $$
  • $$ \color{Red}{i^ \textbf{3}} = i^2 \cdot i = -1 \cdot i = $$
    $$ \color{Red}{ \textbf{-i} } $$
  • $$ \color{Red}{ i^ \textbf{4} } = i^2 \cdot i^2 = -1 \cdot -1 = $$
    $$ \color{Red}{ \textbf{ 1}} $$

Note: Make sure that you understand Table 1 above , because your work with imaginary numbers will continually refer back to it.

What is the larger pattern/formula?

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

  • $$ \color{Red}{i^ \textbf{5}} = i^4 \cdot i = $$
    $$ 1 \cdot i = \color{Red}{ \textbf{ i }} $$
  • $$ \color{Red}{i^ \textbf{6}} = i^4 \cdot i^2 = $$
    $$ 1 \cdot -1= \color{Red}{ \textbf{-1}} $$
  • $$ \color{Red}{ i^ \textbf{7} } = i^4 \cdot i^3 = $$
    $$ 1 \cdot -i= \color{Red}{ \textbf{ -i}} $$
  • $$ \color{Red}{ i^ \textbf{8} } = i^4 \cdot i^4 = $$
    $$ 1 \cdot 1= \color{Red}{ \textbf{ 1}} $$

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

  • $$ \color{Red}{i^ \textbf{9}} = i^4 \cdot i^4 \cdot i = $$
    $$ 1 \cdot 1 \cdot i = \color{Red}{ \textbf{ i }} $$
  • $$ \color{Red}{i^ \textbf{10}} = i^4 \cdot i^4 \cdot i^2 = $$
    $$ 1 \cdot 1 \cdot i^2= \color{Red}{ \textbf{-1}} $$
  • $$ \color{Red}{ i^ \textbf{11} } = i^4 \cdot i^4 \cdot i^3 = $$
    $$ 1 \cdot 1 \cdot -i= \color{Red}{ \textbf{ -i}} $$
  • $$ \color{Red}{ i^ \textbf{12} } = i^4 \cdot i^4 \cdot i^4 = $$
    $$ 1 \cdot 1 \cdot 1= \color{Red}{ \textbf{ 1}} $$

The General Formula

$$ i^k = $$ is the same as $$ i^\color{Red}{r} $$ where $$ \color{Red}{r} $$ is the remainder when k is divided by 4.

Whether the remainder is 1, 2, 3 , or 4 , the key to simplifying powers of i is the remainder when the exponent is divided by 4 . simplifying the powers of the imaginary number i

Practice Problems

Problem 1
$$ i^ {23} $$
Step 1

calculate remainder

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

Step 1

rewrite

$$i^3$$

Step 1

$$ i^{ \color{Red}{3}} = -i $$

Problem 2
$$ i^ {18} $$
Step 1

calculate remainder

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

Step 1

rewrite

$$ i^{ \color{Red}{2}} $$

Step 1

$$ i^{ \color{Red}{2}} = -1 $$

Problem 3
$$ i^ {41} $$
Step 1

calculate remainder

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

Step 1

rewrite

$$ i^{ \color{Red}{1}} $$

Step 1

$$ i^{ \color{Red}{1}} = i $$

Problem 4
$$ i^ {100} $$
Step 1

calculate remainder

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

Step 1

rewrite

$$i^{ \color{Red}{0}} $$

Step 1

$$i^{ \color{Red}{0}} = 1 $$

Problem 5
$$ 5i^ {22} $$

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 remainder

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

Step 1

rewrite

$$i^{ \color{Red}{2}} $$

Step 1

$$i^{ \color{Red}{2}} = -1 $$

Step 1

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

Problem 6
$$ 12i^ {36} $$

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 remainder

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

Step 1

rewrite

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

Step 1

$$ i^{ \color{Red}{0}} = 1 $$

Step 1

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

Problem 6
$$ 7 i^ {103} $$

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 remainder

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

Step 1

rewrite

$$i^{ \color{Red}{3}} $$

Step 1

$$i^{ \color{Red}{3}} = -i $$

Step 1

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