Part I.Simplifying Powers of $$i$$
$$i^5 = ?\\i ^ {21} = ?$$
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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.
$\begin{array}{c|c|c}Expression & Work& Result \\\hline\red{i^ \textbf{9}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^1 = \blue{1} \cdot \blue{1} \cdot i = & \red{ \textbf{ i }}\\\hline\red{i^ \textbf{10}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^2 = \blue{1} \cdot \blue{1} \cdot i^2 = & \red{ \textbf{ -1 }} \\\hline\red{i^ \textbf{11}} & = \blue{i^4} \cdot \blue{i^4} \cdot i^3 = \blue{1} \cdot \blue{1} \cdot i^3 = & \red{ \textbf{ -i }}\\\hline\red{i^ \textbf{12}} & = \blue{i^4} \cdot \blue{i^4} \cdot \blue{i^4} = \blue{1} \cdot \blue{1} \cdot \blue{1}= & \red{ \textbf{ 1 }}\\\hline\end{array}$
The General Formula
$$ i^k$$is the same as $$ i^\red{r} $$ where $$ \red{r} $$ 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 theexponent is divided by 4.
Practice Problems
Problem 1
$$ i^ {23} $$
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
$$ i^ {18} $$
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
$$ i^ {41} $$
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
$$ i^ {100} $$
Step 1
Calculate the remainder
$$ 100 \div 4 $$ has a remainder of $$ \red{0} $$
Step 2
Step 3
Simplify using Table2
$$ i^{ \red{0}} = 1 $$
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 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
Now, the rest
$$ 5 \cdot ( {\color{Blue}-1} ) = -5 $$
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 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
Now, the rest
$$ 12 \cdot ( {\color{Blue} 1} ) = 12 $$
Problem 7
$$ 7 i^ {103} $$
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 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
Now, the rest
$$ 7 \cdot ( {\color{Blue} -i} ) = -7i $$
Worksheet on Imaginary Numbers
Further Reading:
- Complex Numbers Home
- Simplify Negative Radicals
- What is a complex Number
- Complex Numbers Worksheets
