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## Integrated math 2

### Course: Integrated math 2>Unit 5

Lesson 1: The imaginary unit i

# Powers of the imaginary unit

Learn how to simplify any power of the imaginary unit i. For example, simplify i²⁷ as -i.
We know that $i=\sqrt{-1}$ and that ${i}^{2}=-1$.
But what about ${i}^{3}$? ${i}^{4}$? Other integer powers of $i$? How can we evaluate these?

## Finding ${i}^{3}$‍  and ${i}^{4}$‍

The properties of exponents can help us here! In fact, when calculating powers of $i$, we can apply the properties of exponents that we know to be true in the real number system, so long as the exponents are integers.
With this in mind, let's find ${i}^{3}$ and ${i}^{4}$.
We know that ${i}^{3}={i}^{2}\cdot i$. But since ${i}^{2}=-1$, we see that:
$\begin{array}{rl}{i}^{3}& ={i}^{2}\cdot i\\ \\ & =\left(-1\right)\cdot i\\ \\ & =-i\end{array}$
Similarly ${i}^{4}={i}^{2}\cdot {i}^{2}$. Again, using the fact that ${i}^{2}=-1$, we have the following:
$\begin{array}{rl}{i}^{4}& ={i}^{2}\cdot {i}^{2}\\ \\ & =\left(-1\right)\cdot \left(-1\right)\\ \\ & =1\end{array}$

## More powers of $i$‍

Let's keep this going! Let's find the next $4$ powers of $i$ using a similar method.
The results are summarized in the table.
${i}^{1}$${i}^{2}$${i}^{3}$${i}^{4}$${i}^{5}$${i}^{6}$${i}^{7}$${i}^{8}$
$i$$-1$$-i$$1$$i$$-1$$-i$$1$

## An emerging pattern

From the table, it appears that the powers of $i$ cycle through the sequence of $i$, $-1$, $-i$ and $1$.
Using this pattern, can we find ${i}^{20}$? Let's try it!
The following list shows the first $20$ numbers in the repeating sequence.
$i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$
According to this logic, ${i}^{20}$ should be equal to $1$. Let's see if we can support this by using exponents. Remember, we can use the properties of exponents here just like we do with real numbers!
$\begin{array}{rlrl}{i}^{20}& =\left({i}^{4}{\right)}^{5}& & \text{Properties of exponents}\\ \\ & =\left(1{\right)}^{5}& & {i}^{4}=1\\ \\ & =1& & \text{Simplify}\end{array}$
Either way, we see that ${i}^{20}=1$.

## Larger powers of $i$‍

Suppose we now wanted to find ${i}^{138}$. We could list the sequence $i$, $-1$, $-i$, $1$,... out to the ${138}^{\text{th}}$ term, but this would take too much time!
Notice, however, that ${i}^{4}=1$, ${i}^{8}=1$, ${i}^{12}=1$, etc., or, in other words, that $i$ raised to a multiple of $4$ is $1$.
We can use this fact along with the properties of exponents to help us simplify ${i}^{138}$.

### Example

Simplify ${i}^{138}$.

### Solution

While $138$ is not a multiple of $4$, the number $136$ is! Let's use this to help us simplify ${i}^{138}$.
$\begin{array}{rlrl}{i}^{138}& ={i}^{136}\cdot {i}^{2}& & \text{Properties of exponents}\\ \\ & =\left({i}^{4\cdot 34}\right)\cdot {i}^{2}& & 136=4\cdot 34\\ \\ & =\left({i}^{4}{\right)}^{34}\cdot {i}^{2}& & \text{Properties of exponents}\\ \\ & =\left(1{\right)}^{34}\cdot {i}^{2}& & {i}^{4}=1\\ \\ & =1\cdot -1& & {i}^{2}=-1\\ \\ & =-1\end{array}$
So ${i}^{138}=-1$.
Now you might ask why we chose to write ${i}^{138}$ as ${i}^{136}\cdot {i}^{2}$.
Well, if the original exponent is not a multiple of $4$, then finding the closest multiple of $4$ less than it allows us to simplify the power down to $i$, ${i}^{2}$, or ${i}^{3}$ just by using the fact that ${i}^{4}=1$.
This number is easy to find if you divide the original exponent by $4$. It's just the quotient (without the remainder) times $4$.

## Let's practice some problems

### Problem 1

Simplify ${i}^{227}$.

### Problem 2

Simplify ${i}^{2016}$.

### Problem 3

Simplify ${i}^{537}$.

## Challenge Problem

Which of the following is equivalent to ${i}^{-1}$?