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## Integrated math 2

### Course: Integrated math 2>Unit 5

Lesson 1: The imaginary unit i

# 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:
$\begin{array}{rl}{i}^{3}& ={i}^{2}\cdot i\\ \\ & =\left(-1\right)\cdot i\\ \\ & =-i\end{array}$
Similarly ${i}^{4}={i}^{2}\cdot {i}^{2}$. Again, using the fact that ${i}^{2}=-1$, we have the following:
$\begin{array}{rl}{i}^{4}& ={i}^{2}\cdot {i}^{2}\\ \\ & =\left(-1\right)\cdot \left(-1\right)\\ \\ & =1\end{array}$

## 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}^{1}$${i}^{2}$${i}^{3}$${i}^{4}$${i}^{5}$${i}^{6}$${i}^{7}$${i}^{8}$
$i$$-1$$-i$$1$$i$$-1$$-i$$1$

## 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.
$i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$, $i$, $-1$, $-i$, $1$
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!
$\begin{array}{rlrl}{i}^{20}& =\left({i}^{4}{\right)}^{5}& & \text{Properties of exponents}\\ \\ & =\left(1{\right)}^{5}& & {i}^{4}=1\\ \\ & =1& & \text{Simplify}\end{array}$
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 multiple of $4$ is $1$.
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 ${i}^{138}$.
$\begin{array}{rlrl}{i}^{138}& ={i}^{136}\cdot {i}^{2}& & \text{Properties of exponents}\\ \\ & =\left({i}^{4\cdot 34}\right)\cdot {i}^{2}& & 136=4\cdot 34\\ \\ & =\left({i}^{4}{\right)}^{34}\cdot {i}^{2}& & \text{Properties of exponents}\\ \\ & =\left(1{\right)}^{34}\cdot {i}^{2}& & {i}^{4}=1\\ \\ & =1\cdot -1& & {i}^{2}=-1\\ \\ & =-1\end{array}$
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

Simplify ${i}^{227}$.

### Problem 2

Simplify ${i}^{2016}$.

### Problem 3

Simplify ${i}^{537}$.

## Challenge Problem

Which of the following is equivalent to ${i}^{-1}$?

## 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
• 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.
• On hour 9 of study. send help. xD
help sent
• 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

Thanks so much!!
• 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
• 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.
• how is i^-1 = -i
doesnt it equal 1/i
• 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.
• where are the powers of the imaginary unit? I've seen no powers. Does it throw fireballs?
• Is there any reason to use multiple of 4 over multiple of 1,2 etc or is just arbitrary?
• 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.
• what Will be i to the power i 899999
• i^899999
(i^900000)/i
1/i
i/(i^2)
i/(-1)
-i
• 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