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## Algebra 2

### Course: Algebra 2 > Unit 2

Lesson 1: The imaginary unit i- Intro to the imaginary numbers
- Intro to the imaginary numbers
- Simplifying roots of negative numbers
- Simplify roots of negative numbers
- Powers of the imaginary unit
- Powers of the imaginary unit
- Powers of the imaginary unit
- i as the principal root of -1

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# Powers of the imaginary unit

CCSS.Math: , ,

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:

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:

## More powers of i

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

The results are summarized in the table.

i, start superscript, 1, end superscript | i, squared | i, cubed | i, start superscript, 4, end superscript | i, start superscript, 5, end superscript | i, start superscript, 6, end superscript | i, start superscript, 7, end superscript | i, start superscript, 8, end superscript |
---|---|---|---|---|---|---|---|

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 |

## 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!

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

*is 1.***multiple of 4**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.

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

### Problem 2

### Problem 3

## Challenge Problem

## Want to join the conversation?

- How is
`i^3 = -i`

?

My understanding leads me to:`i^3 = i * i * i`

i^3 = sqrt( -1 ) * sqrt( -1 ) * sqrt( -1 ) ; i = sqrt( -1 )

i^3 = sqrt( -1 * -1 * -1 )

i^3 = sqrt( -1 )

i^3 = i(47 votes)- On step 3 you did:

√𝑎 • √𝑏 = √(𝑎𝑏)

But that is only true if:

𝑎, 𝑏 > 0

Which is not the case here. Comment if you want to see a proof of that.(132 votes)

- Hi, could someone maybe help me with what I did wrong in the last challenge problem? I followed along with their explanation well enough, but I'm scratching my head as to exactly where I messed up the problem. My answer was i, while the correct was -i. Also, by way of explanation, in steps three, four, and five, I am assuming that 1 can be rewritten as 1=sqrt(1) because 1*1= 1, therefore the square root of 1 equals 1.

i^-1=

1/i^1 (because properties of negative exponents)

1/i (because i^1=i)

1/sqrt(-1) (because i=sqrt(-1))

sqrt(1)/sqrt(-1) (because the 1 in the numerator can be rewritten as sqrt(1) without changing its value)

sqrt(1/-1) (to simplify inside the square root)

sqrt(-1) (because 1/-1 is -1)

sqrt(-1)=i (because i=sqrt(-1))

=i

It's driving me crazy that I can't figure out what I did wrong!! Please help.

Thanks so much!!(13 votes)- It seems to me your problem is the definition
`i = sqrt(-1)`

. Better use`i^2 = -1`

. You can see, that your last step would get you`sqrt(-1) = +/- i`

, so the right solution is at least within, but the problem is better solved another way:`i ^ (-1) = 1/i = 1/i * i/i`

<-- Expand with i`= i/(i^2)`

<-- Multiply`= i/(-1)`

<-- Definition`= -i`

<-- Simplify

Therefore`i^(-1) = -i`

(26 votes)

- Where do you use this in the real world?(13 votes)
- 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).(21 votes)

- Fast way to solve this: Divide the power by 4 and then GET THE REMAINDER

Ik u probaby never used this past 4th grade

what the remainder number is is how many times you count after the last tern in the sequence, which is 1.

Upvote pls i neec clout(14 votes) - 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.(4 votes)

- how is i^-1 = -i

doesnt it equal 1/i(8 votes)- Yes, i^-1 = 1/i.

Notice that you can multiply 1/i by i/i, which gives you i/-1, or -i, which is much easier to comprehend than 1/i.(12 votes)

- what Will be i to the power i 899999(5 votes)
- i^899999

(i^900000)/i

1/i

i/(i^2)

i/(-1)

-i(10 votes)

- what is the answer of i ^ i ?(7 votes)
- 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.(5 votes)

- where are the powers of the imaginary unit? I've seen no powers. Does it throw fireballs?(8 votes)
- if i = √-1

i^2 = √-1 * √-1

= √(-1 * -1) <-- Why this doesn't work?

= √1

= 1(5 votes)- 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.(6 votes)