Any even power of i..........will always be real................and it will always be either a neg 1.......or 1...............


Powers of the imaginary unit

Learn how to simplify any power of the imaginary unit i. For example, simplify i²⁷ as -i.

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

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

Finding $i^3$i, start superscript, 3, end superscript and $i^4$i, start superscript, 4, end superscript

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, start superscript, 3, end superscript and $i^4$i, start superscript, 4, end superscript.

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

$\begin{aligned} i^3 &= {{i^2}}\cdot i\\ \\& ={ (-1)}\cdot i\\ \\& = \purpleD{-i} \end{aligned}$

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

$\begin{aligned} i^4 &= {{i^2\cdot i^2}}\\ \\& =({ -1})\cdot ({-1})\\ \\& = \goldD{1} \end{aligned}$

More powers of $i$i

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

i5=i4⋅i     =1⋅i=iProperties of exponentsSince i4=1

i6=i4⋅i2=1⋅(−1)=−1Properties of exponentsSince i4=1 and i2=−1

i7=i4⋅i3=1⋅(−i)=−iProperties of exponentsSince i4=1 and i3=−i

i8=i4⋅i4    =1⋅1=1Properties of exponentsSince i4=1 

The results are summarized in the table.

$i^1$i, start superscript, 1, end superscript	$i^2$i, start superscript, 2, end superscript	$i^3$i, start superscript, 3, end superscript	$i^4$i, start superscript, 4, end superscript	$i^5$i, start superscript, 5, end superscript	$i^6$i, start superscript, 6, end superscript	$i^7$i, start superscript, 7, end superscript	$i^8$i, start superscript, 8, end superscript
$\blueD i$start color blueD, i, end color blueD	$\greenD{-1}$start color greenD, minus, 1, end color greenD	$\purpleD{-i}$start color purpleD, minus, i, end color purpleD	$\goldD 1$start color goldD, 1, end color goldD	$\blueD i$start color blueD, i, end color blueD	$\greenD{-1}$start color greenD, minus, 1, end color greenD	$\purpleD{-i}$start color purpleD, minus, i, end color purpleD	$\goldD 1$start color goldD, 1, end color goldD

An emerging pattern

From the table, it appears that the powers of $i$i cycle through the sequence of $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD and $\goldD1$start color goldD, 1, end color goldD.

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.

$\quad$space$\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD, $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD, $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD, $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD, $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD

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

Larger powers of $i$i

Suppose we now wanted to find $i^{138}$i, start superscript, 138, end superscript. We could list the sequence $\blueD i$start color blueD, i, end color blueD, $\greenD{-1}$start color greenD, minus, 1, end color greenD, $\purpleD{-i}$start color purpleD, minus, i, end color purpleD, $\goldD 1$start color goldD, 1, end color goldD,... out to the $138^\text{th}$138, start superscript, t, h, 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.

Example

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

Solution

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.

i138=i136⋅i2=(i4⋅34)⋅i2=(i4)34⋅i2=(1)34⋅i2=1⋅−1=−1Properties of exponents136=4⋅34Properties of exponentsi4=1i2=−1

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, start superscript, 2, end superscript.

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, start superscript, 2, end superscript, or $i^3$i, start superscript, 3, end superscript just by using the fact that $i^4=1$i, start superscript, 4, end superscript, equals, 1.