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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, equals, square root of, minus, 1, end square root and that i, squared, equals, minus, 1.
But what about i, cubed? i, start superscript, 4, end superscript? Other integer powers of i? How can we evaluate these?

Finding i, cubed and i, start superscript, 4, end superscript

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, cubed and i, start superscript, 4, end superscript.
We know that i, cubed, equals, i, squared, dot, i. But since i, squared, equals, minus, 1, we see that:
i3=i2i=(1)i=i\begin{aligned} i^3 &= {{i^2}}\cdot i\\ \\ &={ (-1)}\cdot i\\ \\ &= \purpleD{-i} \end{aligned}
Similarly i, start superscript, 4, end superscript, equals, i, squared, dot, i, squared. Again, using the fact that i, squared, equals, minus, 1, we have the following:
i4=i2i2=(1)(1)=1\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.
i5=i4iProperties of exponents=1iSince i4=1=ii6=i4i2Properties of exponents=1(1)Since i4=1 and i2=1=1i7=i4i3Properties of exponents=1(i)Since i4=1 and i3=i=ii8=i4i4Properties of exponents=11Since i4=1=1\begin{aligned} i^5 &= {i^4\cdot i}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot i&&{\gray{\text{Since }i^4=1}} \\\\ &= \blueD i \\\\ \\\\ i^6 &= {i^4\cdot i^2}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot (-1)&&{\gray{\text{Since }i^4=1\text{ and }i^2=-1}} \\\\ &=\greenD{-1} \\\\ \\\\ i^7 &= {i^4\cdot i^3}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot (-i)&&{\gray{\text{Since }i^4=1\text{ and }i^3=-i}} \\\\ &=\purpleD{-i} \\\\ \\\\ i^8 &= {i^4\cdot i^4}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot 1&&{\gray{\text{Since }i^4=1}} \\\\ &=\goldD 1 \end{aligned}
The results are summarized in the table.
i, start superscript, 1, end superscripti, squaredi, cubedi, start superscript, 4, end superscripti, start superscript, 5, end superscripti, start superscript, 6, end superscripti, start superscript, 7, end superscripti, start superscript, 8, end superscript
start color #11accd, i, end color #11accdstart color #1fab54, minus, 1, end color #1fab54start color #7854ab, minus, i, end color #7854abstart color #e07d10, 1, end color #e07d10start color #11accd, i, end color #11accdstart color #1fab54, minus, 1, end color #1fab54start color #7854ab, minus, i, end color #7854abstart color #e07d10, 1, end color #e07d10

An emerging pattern

From the table, it appears that the powers of i cycle through the sequence of start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab and start color #e07d10, 1, end color #e07d10.
Using this pattern, can we find i, start superscript, 20, end superscript? Let's try it!
The following list shows the first 20 numbers in the repeating sequence.
start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10, start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10, start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10, start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10, start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10
According to this logic, i, start superscript, 20, end superscript should be equal to 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!
i20=(i4)5Properties of exponents=(1)5i4=1=1Simplify\begin{aligned} i^{20} &= (i^4)^5&&{\gray{\text{Properties of exponents}}} \\\\ &= (1)^5 &&{\gray{i^4=1}} \\\\ &= \goldD 1 &&{\gray{\text{Simplify}}} \end{aligned}
Either way, we see that i, start superscript, 20, end superscript, equals, 1.

Larger powers of i

Suppose we now wanted to find i, start superscript, 138, end superscript. We could list the sequence start color #11accd, i, end color #11accd, start color #1fab54, minus, 1, end color #1fab54, start color #7854ab, minus, i, end color #7854ab, start color #e07d10, 1, end color #e07d10,... out to the 138, start superscript, start text, t, h, end text, end superscript term, but this would take too much time!
Notice, however, that i, start superscript, 4, end superscript, equals, 1, i, start superscript, 8, end superscript, equals, 1, i, start superscript, 12, end superscript, equals, 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, start superscript, 138, end superscript.

Example

Simplify i, start superscript, 138, end superscript.

Solution

While 138 is not a multiple of 4, the number 136 is! Let's use this to help us simplify i, start superscript, 138, end superscript.
i138=i136i2Properties of exponents=(i434)i2136=434=(i4)34i2Properties of exponents=(1)34i2i4=1=11i2=1=1\begin{aligned} i^{138}&=i^{136}\cdot i^2&&{\gray{\text{Properties of exponents}}} \\\\ &=(i^{4\cdot 34})\cdot i^2&&{\gray{136=4\cdot 34}} \\\\ &=(i^{4})^{34}\cdot i^2&&{\gray{\text{Properties of exponents}}} \\\\ &=(1)^{34}\cdot i^2 &&{\gray{i^4=1}} \\\\ &=1\cdot -1&&{\gray{i^2=-1}} \\\\ &=-1 \end{aligned}
So i, start superscript, 138, end superscript, equals, minus, 1.
Now you might ask why we chose to write i, start superscript, 138, end superscript as i, start superscript, 136, end superscript, dot, i, squared.
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, squared, or i, cubed just by using the fact that i, start superscript, 4, end superscript, equals, 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, start superscript, 227, end superscript.

Problem 2

Simplify i, start superscript, 2016, end superscript.

Problem 3

Simplify i, start superscript, 537, end superscript.

Challenge Problem

Which of the following is equivalent to i, start superscript, minus, 1, end superscript?
Choose 1 answer:
Choose 1 answer:

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