Powers of the imaginary unit

We know that $i=\sqrt{-1}$i, equals, square root of, minus, 1, end square root and that $i^2=-1$i, squared, equals, minus, 1.

But what about $i^3$i, cubed? $i^4$i, start superscript, 4, end superscript? Other integer powers of $i$i? How can we evaluate these?

The properties of exponents can help us here! In fact, when calculating powers of $i$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 $i^3$i, cubed and $i^4$i, start superscript, 4, end superscript.

We know that $i^3=i^2\cdot i$i, cubed, equals, i, squared, dot, i. But since ${i^2=-1}$i, squared, equals, minus, 1, we see that:

Similarly $i^4=i^2\cdot i^2$i, start superscript, 4, end superscript, equals, i, squared, dot, i, squared. Again, using the fact that ${i^2=-1}$i, squared, equals, minus, 1, we have the following:

Let's keep this going! Let's find the next $4$4 powers of $i$i using a similar method.

The results are summarized in the table.

From the table, it appears that the powers of $i$i cycle through the sequence of $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab and $\goldD1$start color #e07d10, 1, end color #e07d10.

Using this pattern, can we find $i^{20}$i, start superscript, 20, end superscript? Let's try it!

The following list shows the first $20$20 numbers in the repeating sequence.

$\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10, $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10, $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10, $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10, $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10

According to this logic, $i^{20}$i, start superscript, 20, end superscript should be equal to $\goldD 1$start color #e07d10, 1, end color #e07d10. 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!

Either way, we see that $i^{20}=1$i, start superscript, 20, end superscript, equals, 1.

Suppose we now wanted to find $i^{138}$i, start superscript, 138, end superscript. We could list the sequence $\blueD i$start color #11accd, i, end color #11accd, $\greenD{-1}$start color #1fab54, minus, 1, end color #1fab54, $\purpleD{-i}$start color #7854ab, minus, i, end color #7854ab, $\goldD 1$start color #e07d10, 1, end color #e07d10,... out to the $138^\text{th}$138, start superscript, start text, t, h, end text, end superscript term, but this would take too much time!

Notice, however, that $i^4=1$i, start superscript, 4, end superscript, equals, 1, $i^8=1$i, start superscript, 8, end superscript, equals, 1, $i^{12}=1$i, start superscript, 12, end superscript, equals, 1, etc., or, in other words, that $i$i raised to a multiple of $4$4 is $1$1.

We can use this fact along with the properties of exponents to help us simplify $i^{138}$i, start superscript, 138, end superscript.

Simplify $i^{138}$i, start superscript, 138, end superscript.

While $138$138 is not a multiple of $4$4, the number $136$136 is! Let's use this to help us simplify $i^{138}$i, start superscript, 138, end superscript.

So $i^{138}=-1$i, start superscript, 138, end superscript, equals, minus, 1.

Now you might ask why we chose to write $i^{138}$i, start superscript, 138, end superscript as $i^{136}\cdot i^2$i, start superscript, 136, end superscript, dot, i, squared.

Well, if the original exponent is not a multiple of $4$4, then finding the closest multiple of $4$4 less than it allows us to simplify the power down to $i$i, $i^2$i, squared, or $i^3$i, cubed just by using the fact that $i^4=1$i, start superscript, 4, end superscript, equals, 1.

This number is easy to find if you divide the original exponent by $4$4. It's just the quotient (without the remainder) times $4$4.