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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 $i^3$i, cubed and $i^4$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:
\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:
\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 powers of i using a similar method.
\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!
\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 $i$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.
\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?

## Want to join the conversation?

• How is i^3 = -i ?
i^3 = i * i * ii^3 = sqrt( -1 ) * sqrt( -1 ) * sqrt( -1 )    ; i = sqrt( -1 )i^3 = sqrt( -1 * -1 * -1 )i^3 = sqrt( -1 )i^3 = i
• It's better to use the definition i^2 = -1.
Therefore i^3 = i^2 * i = -i. While working with imaginary Numbers you have to be careful about the definition you choose.
• Where do you use this in the real world?
• Physics uses it. Electrical Engineering often have to use the imaginary unit in their calculations, but it is also used in Robotics. There, to rotate an object through 3 dimensions they even result in using Quaternions (which build on the imaginary unit i).
• will i^0=1?
• Hey, the explanation to your query is "any number other than 0 taken to the power 0 is defined to be 1".
So i=√-1, "i" here is a real number. So i^0=1.
It is explained in great detail at this website-->
(though it requires some math, though eventually, you get the idea)
http://mathworld.wolfram.com/Power.html
However, "zero to the power of zero" (or 0^0) is undefined.
• what Will be i to the power i 899999
• i^899999
(i^900000)/i
1/i
i/(i^2)
i/(-1)
-i
• what is the answer of i ^ i ?
• Well, due to a property that you might learn later, i = e^(i*pi/2). So, i^i = e^(i*pi/2*i). We know i*i = -1, so i^i = e^(-pi/2).

i^i is always a real number, but there are some problems with this definition based on the property.
• if i = √-1

i^2 = √-1 * √-1
= √(-1 * -1) <-- Why this doesn't work?
= √1
= 1
• The property √a√b=√(ab) doesn't hold if both a and b are both negative. What you've written here is a proof of that.
• how to remember the pattern for i raised to a power of a number
• Here's how I do it visually.
Refer to this: https://en.wikipedia.org/wiki/Imaginary_unit

Look at the picture to the right. That's the imaginary number unit circle. It's significance is not needed to know as of right now, but the main thing is it has all four variants.

i⁰ = 1
i¹ = i
i² = -1
i³= -i
i⁴ = 1

You may notice that on the unit circle, the value is going counterclockwise. Each multiple of four will always equal 1.

i¹⁶ = 1
i²⁴ = 1

Always refer to the multiple of four, as if you know that, if you have (for example) i¹⁰¹:

We can split it up by a multiple of 4 times some other number:
i¹⁰⁰ * i¹
100 is a multiple of 4, so it equals 1
1 * i = i

hopefully this helps !
• I don't understand how the last property is being explained. It says the property can be verified algebraically. Specifically I'm not sure why i^3 or i^4 is being used in this case. What do they have to do with what 1/i is?
(1 vote)
• i^3=i^4*i^-1

Why?
We know that
a^n*a^m=a^(n+m)
so
i^4*i^-1
is actually
i^(4-1)=i^3
That's why.
If you wanted to know why
i^-1=-i
here is the explaination: Because
i^4*i^-1=i^3
and
i^3=-i
also
i^4=1
because
-----------------------|i^1| |i^2| |i^3| |i^4|| i | |-1 | |-i | | 1 |-----------------------
(the pattern written in the document). We can simplify it:
i^4*i^-1=i^3 1 *i^-1=-1    i^-1=-1
(1 vote)
• Hi, I am a 5th grader at Bandelier Elementary School in Albuquerque, NM. I am learning precalculus. My brother is in his second year of High School, and he hasn't started learning pre-calculus yet. I'm ahead!