# Powers of the imaginary unit

CCSS Math: HSN.CN.A.2
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{aligned} i^3 &= {{i^2}}\cdot i\\ \\ &={ (-1)}\cdot i\\ \\ &= \purpleD{-i} \end{aligned}
Similarly $i^4=i^2\cdot i^2$. Again, using the fact that ${i^2=-1}$, we have the following:
\begin{aligned} i^4 &= {{i^2\cdot i^2}}\\ \\ &=({ -1})\cdot ({-1})\\ \\ &= \goldD{1} \end{aligned}

## More powers of $i$

Let's keep this going! Let's find the next $4$ powers of $i$ using a similar method.
\begin{aligned} \Large i^5 &= {i^4\cdot i}~~~~~&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=1\cdot i&&\small{\gray{\text{Since i^4=1}}}\\ \\ &= \blueD i \end{aligned}
\begin{aligned}\Large i^6 &= {i^4\cdot i^2}&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=1\cdot (-1)&&\small{\gray{\text{Since i^4=1 and i^2=-1}}}\\ \\ &=\greenD{-1} \end{aligned}
\begin{aligned}\Large i^7 &= {i^4\cdot i^3}&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=1\cdot (-i)&&\small{\gray{\text{Since i^4=1 and i^3=-i}}}\\ \\ &=\purpleD{-i} \end{aligned}
\begin{aligned}\Large i^8 &= {i^4\cdot i^4~~~~}&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=1\cdot 1&&\small{\gray{\text{Since i^4=1 }}}\\ \\ &=\goldD 1 \end{aligned}
The results are summarized in the table.
$i^1$$i^2$$i^3$$i^4$$i^5$$i^6$$i^7$$i^8$
$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$

## An emerging pattern

From the table, it appears that the powers of $i$ cycle through the sequence of $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$ and $\goldD1$.
Using this pattern, can we find $i^{20}$? Let's try it!
The following list shows the first $20$ numbers in the repeating sequence.
$\quad$$\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$
According to this logic, $i^{20}$ should be equal to $\goldD 1$. 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{aligned} i^{20} &= (i^4)^5&&\small{\gray{\text{Properties of exponents}}}\\ \\ &= (1)^5 &&\small{\gray{i^4=1}}\\\\ &= \goldD 1 &&\small{\gray{\text{Simplify}}}\end{aligned}
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 $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$,... 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{aligned} i^{138} &=i^{136}\cdot i^2 &&\small{\gray{\text{Properties of exponents}}}\\\\ &=(i^{4\cdot 34})\cdot i^2&&\small{\gray{136=4\cdot 34}} \\\\ &=(i^{4})^{34}\cdot i^2&&\small{\gray{\text{Properties of exponents}}} \\\\ &=(1)^{34}\cdot i^2 &&\small{\gray{\text{i^4=1}}}\\\\ &=1\cdot -1&&\small{\gray{\text{i^2=-1}}}\\\\ &=-1 \end{aligned}
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 $4$.
Sure. Let's look at the exponent in this particular example: $138$.
Below is the long division for $138\div 4$.
\qquad\begin{aligned} \\ \\ 34& \\ {4}~\overline{)~138}& \\ \underline{12~~}& \\ 18& \\ \underline{16}& \\ 2& \\ & \\ & \\ \end{aligned}
Notice that $34\cdot 4 = 136$. This is the closest multiple of $4$ that is less than $138$.
Sure! Watch this video to see Sal simplify several powers of $i$.

## Let's practice some problems

### Problem 1

Simplify $i^{227}$.
The closest multiple of $4$ less than $227$ is $224$.
Let's use this to simplify $i^{227}$ as follows:
\begin{aligned} i^{227} &= i^{224}\cdot i^3&&\small{\gray{\text{Properties of exponents}}} \\\\&= (i^{4})^{56}\cdot i^3&&\small{\gray{\text{Properties of exponents}}} \\\\ &=(1)^{56}\cdot i^3 &&\small{\gray{\text{i^4=1}}}\\\\ &=1\cdot -i&&\small{\gray{\text{i^3=-i}}}\\\\ &=-i \end{aligned}
So $i^{227}=-i$.

### Problem 2

Simplify $i^{2016}$.
The number $2016$ is a multiple of $4$, and so the answer is $1$. The work below shows a bit more detail if needed.
\begin{aligned} i^{2016} &= (i^{4})^{504}&&\small{\gray{\text{Properties of exponents}}} \\\\ &=(1)^{504} &&\small{\gray{\text{i^4=1}}}\\\\ &=1&&\small{\gray{\text{Simplify}}}\\\\ \end{aligned}

### Problem 3

Simplify $i^{537}$.
The closest multiple of $4$ less than $537$ is $536$.
Let's use this to simplify $i^{537}$ as follows:
\begin{aligned} i^{537} &= i^{536}\cdot i&&\small{\gray{\text{Exponent properties}}} \\\\&= (i^{4})^{134}\cdot i&&\small{\gray{\text{Exponent properties}}} \\\\ &=(1)^{134}\cdot i &&\small{\gray i^4=1}\\\\ &=1\cdot i\\\\ &=i \end{aligned}
So $i^{537}=i$.

## Challenge Problem

Which of the following is equivalent to $i^{-1}$?
$i^1$$i^2$$i^3$$i^4$$i^5$$i^6$$i^7$$i^8$
$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$
However, instead of extending the pattern to the right (for exponents $>1$), we can think about extending the pattern to the left (for exponents $<1$). In fact, since we want to know $i^{-1}$, we can just extend the sequence two places to the left.
$i^{-1}$$i^0$$i^1$$i^2$$i^3$$i^4$$i^5$$i^6$$i^7$$i^8$
$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$$\blueD i$$\greenD{-1}$$\purpleD{-i}$$\goldD 1$
$i^{-1}$$i^0$$i^1$$i^2$$i^3$$i^4$$i^5$$i^6$$i^7$$i^8$
$\purpleD{-i}$$\goldD1$$\gray i$$\gray{-1}$$\gray{-i}$$\gray 1$$\gray i$$\gray{-1}$$\gray{-i}$$\gray 1$
So $i^{-1}=-i$.
\begin{aligned}i^3&=i^4\cdot i^{-1}\\\\ -i&=1\cdot i^{-1}&\small{\gray{\text{Since }i^3=-i \text{ and } i^4=1}}\\\\ -i&=i^{-1} \end{aligned}