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=1i=\sqrt{-1} and that i2=1i^2=-1.
But what about i3i^3? i4i^4? Other integer powers of ii? How can we evaluate these?

Finding i3i^3 and i4i^4

The properties of exponents can help us here! In fact, when calculating powers of ii, 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 i3i^3 and i4i^4.
We know that i3=i2ii^3=i^2\cdot i. But since i2=1{i^2=-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 i4=i2i2i^4=i^2\cdot i^2. Again, using the fact that i2=1{i^2=-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 ii

Let's keep this going! Let's find the next 44 powers of ii using a similar method.
i5=i4i     Properties of exponents=1iSince i4=1=i\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}
i6=i4i2Properties of exponents=1(1)Since  and i4=1i2=1=1\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}
i7=i4i3Properties of exponents=1(i)Since  and i4=1i3=i=i\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}
i8=i4i4    Properties of exponents=11Since  i4=1=1\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.
i1i^1i2i^2i3i^3i4i^4i5i^5i6i^6i7i^7i8i^8
i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1

An emerging pattern

From the table, it appears that the powers of ii cycle through the sequence of i\blueD i, 1\greenD{-1}, i\purpleD{-i} and 1\goldD1.
Using this pattern, can we find i20i^{20}? Let's try it!
The following list shows the first 2020 numbers in the repeating sequence.
\quadi\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1, i\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1, i\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1, i\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1, i\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1
According to this logic, i20i^{20} should be equal to 1\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!
i20=(i4)5Properties of exponents=(1)5i4=1=1Simplify\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 i20=1i^{20}=1.

Larger powers of ii

Suppose we now wanted to find i138i^{138}. We could list the sequence i\blueD i, 1\greenD{-1}, i\purpleD{-i}, 1\goldD 1,... out to the 138th138^\text{th} term, but this would take too much time!
Notice, however, that i4=1i^4=1, i8=1i^8=1, i12=1i^{12}=1, etc., or, in other words, that ii raised to a multiple of 44 is 11.
We can use this fact along with the properties of exponents to help us simplify i138i^{138}.

Example

Simplify i138i^{138}.

Solution

While 138138 is not a multiple of 44, the number 136136 is! Let's use this to help us simplify i138i^{138}.
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 &&\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 i138=1i^{138}=-1.
Now you might ask why we chose to write i138i^{138} as i136i2i^{136}\cdot i^2.
Well, if the original exponent is not a multiple of 44, then finding the closest multiple of 44 less than it allows us to simplify the power down to ii, i2i^2, or i3i^3 just by using the fact that i4=1i^4=1.
This number is easy to find if you divide the original exponent by 44. It's just the quotient (without the remainder) times 44.
Sure. Let's look at the exponent in this particular example: 138138.
Below is the long division for 138÷4138\div 4.
344 ) 13812  18162\qquad\begin{aligned} \\ \\ 34& \\ {4}~\overline{)~138}& \\ \underline{12~~}& \\ 18& \\ \underline{16}& \\ 2& \\ & \\ & \\ \end{aligned}
Notice that 344=13634\cdot 4 = 136. This is the closest multiple of 44 that is less than 138138.
Sure! Watch this video to see Sal simplify several powers of ii.

Let's practice some problems

Problem 1

Simplify i227i^{227}.
The closest multiple of 44 less than 227227 is 224224.
Let's use this to simplify i227i^{227} as follows:
i227=i224i3Properties of exponents=(i4)56i3Properties of exponents=(1)56i3i4=1=1ii3=i=i\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 i227=ii^{227}=-i.

Problem 2

Simplify i2016i^{2016}.
The number 20162016 is a multiple of 44, and so the answer is 11. The work below shows a bit more detail if needed.
i2016=(i4)504Properties of exponents=(1)504i4=1=1Simplify\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 i537i^{537}.
The closest multiple of 44 less than 537537 is 536536.
Let's use this to simplify i537i^{537} as follows:
i537=i536iExponent properties=(i4)134iExponent properties=(1)134ii4=1=1i=i\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 i537=ii^{537}=i.

Challenge Problem

Which of the following is equivalent to i1i^{-1}?
Choose 1 answer:
Choose 1 answer:
One easy way to approach this problem is to use and extend upon the pattern.
i1i^1i2i^2i3i^3i4i^4i5i^5i6i^6i7i^7i8i^8
i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1
However, instead of extending the pattern to the right (for exponents >1>1), we can think about extending the pattern to the left (for exponents <1<1). In fact, since we want to know i1i^{-1}, we can just extend the sequence two places to the left.
i1i^{-1}i0i^0i1i^1i2i^2i3i^3i4i^4i5i^5i6i^6i7i^7i8i^8
i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1i\blueD i1\greenD{-1}i\purpleD{-i}1\goldD 1
Filling in the blanks, we have:
i1i^{-1}i0i^0i1i^1i2i^2i3i^3i4i^4i5i^5i6i^6i7i^7i8i^8
i\purpleD{-i}1\goldD1i\gray i1\gray{-1}i\gray{-i}1\gray 1i\gray i1\gray{-1}i\gray{-i}1\gray 1
So i1=ii^{-1}=-i.
We can also verify this algebraically by using properties of exponents as shown below:
i3=i4i1i=1i1Since i3=i and i4=1i=i1\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}
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