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### Course: Algebra (all content) > Unit 16

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

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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 $i=\sqrt{-1}$ and that ${i}^{2}=-1$ .

But what about ${i}^{3}$ ? ${i}^{4}$ ? Other integer powers of $i$ ? How can we evaluate these?

## Finding ${i}^{3}$ and ${i}^{4}$

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 ${i}^{3}$ and ${i}^{4}$ .

We know that ${i}^{3}={i}^{2}\cdot i$ . But since ${i}^{2}=-1$ , we see that:

Similarly ${i}^{4}={i}^{2}\cdot {i}^{2}$ . Again, using the fact that ${i}^{2}=-1$ , 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.

## An emerging pattern

From the table, it appears that the powers of $i$ cycle through the sequence of ${i}$ , ${-1}$ , ${-i}$ and ${1}$ .

Using this pattern, can we find ${i}^{20}$ ? Let's try it!

The following list shows the first $20$ numbers in the repeating sequence.

According to this logic, ${i}^{20}$ should be equal to ${1}$ . 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!

Either way, we see that ${i}^{20}=1$ .

## Larger powers of $i$

Suppose we now wanted to find ${i}^{138}$ . We could list the sequence ${i}$ , ${-1}$ , ${-i}$ , ${1}$ ,... out to the ${138}^{\text{th}}$ term, but this would take too much time!

Notice, however, that ${i}^{4}=1$ , ${i}^{8}=1$ , ${i}^{12}=1$ , etc., or, in other words, that $i$ raised to a $1$ .

*is***multiple of**$4$ We can use this fact along with the properties of exponents to help us simplify ${i}^{138}$ .

### Example

Simplify ${i}^{138}$ .

### Solution

While $138$ is not a multiple of $4$ , the number $136$ is! Let's use this to help us simplify ${i}^{138}$ .

So ${i}^{138}=-1$ .

Now you might ask why we chose to write ${i}^{138}$ as ${i}^{136}\cdot {i}^{2}$ .

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 $i$ , ${i}^{2}$ , or ${i}^{3}$ just by using the fact that ${i}^{4}=1$ .

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(53 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.(154 votes)

- On hour 9 of study. send help. xD(41 votes)
- sending help your way
*help sent*(21 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!!(14 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`

(32 votes)

- Where do you use this in the real world?(14 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).(29 votes)

- how is i^-1 = -i

doesnt it equal 1/i(7 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.(18 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.(6 votes)

- where are the powers of the imaginary unit? I've seen no powers. Does it throw fireballs?(13 votes)
- what Will be i to the power i 899999(4 votes)
- i^899999

(i^900000)/i

1/i

i/(i^2)

i/(-1)

-i(12 votes)

- I understand how to solve for i using simple negative exponents, but I don't know how you would solve for more larger negative exponents.(1 vote)
- I’ll walk you through an example.

Say you want to evaluate i^-207**1. Divide by four and find the remainder**

207/4 = 51 R 3

Since the remainder is 3, i^-207 = i^-3**To see why this works: Using exponent properties, you can rewrite the problem as**

(i^4)^-51 * i^-3

=1^-51 * i^-3

=1 * i^-3

=i^-3**2. Now, this is something you know how to solve**

i^-3

=1/i^3

=1/-i

=1/-i * i/i

=i/-i^2

=i/-(-1)

=i/1

=i.

i^-207 = i(18 votes)

- Is there any reason to use multiple of 4 over multiple of 1,2 etc or is just arbitrary?(4 votes)
- It must be 4. It is not arbitrary. The powers of i rotate thru 4 values, not something else.

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

Then this pattern repeats

i^5 = i

i^6 = -1

i^7 = -i

i^8 = 1

etc.

Hope this helps.(10 votes)