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 ```

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

sending help your way _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 It's driving me crazy that I can't figure out what I did wrong!! Please help. 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).

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.

what Will be i to the power i 899999

i^899999 (i^900000)/i 1/i i/(i^2) i/(-1) -i

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.

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.

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

Main content

Course: Algebra 2 > Unit 2

Lesson 1: The imaginary unit i

Powers of the imaginary unit

Q: 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).

Q: 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.

Google Classroom

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{aligned} i^{3} & = i^{2} \cdot i \\ = (- 1) \cdot i \\ = - i \end{aligned}

Similarly

i^{4} = i^{2} \cdot i^{2}

. Again, using the fact that

i^{2} = - 1

, we have the following:

\begin{aligned} i^{4} & = i^{2} \cdot i^{2} \\ = (- 1) \cdot (- 1) \\ = 1 \end{aligned}

More powers of $i$ ‍

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 & Properties of exponents \\ = 1 \cdot i & Since i^{4} = 1 \\ = i \\ i^{6} & = i^{4} \cdot i^{2} & Properties of exponents \\ = 1 \cdot (- 1) & Since i^{4} = 1 and i^{2} = - 1 \\ = - 1 \\ i^{7} & = i^{4} \cdot i^{3} & Properties of exponents \\ = 1 \cdot (- i) & Since i^{4} = 1 and i^{3} = - i \\ = - i \\ i^{8} & = i^{4} \cdot i^{4} & Properties of exponents \\ = 1 \cdot 1 & Since i^{4} = 1 \\ = 1 \end{aligned}

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{aligned} i^{20} & = (i^{4})^{5} & Properties of exponents \\ = (1)^{5} & i^{4} = 1 \\ = 1 & Simplify \end{aligned}

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^{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{aligned} i^{138} & = i^{136} \cdot i^{2} & Properties of exponents \\ = (i^{4 \cdot 34}) \cdot i^{2} & 136 = 4 \cdot 34 \\ = (i^{4})^{34} \cdot i^{2} & Properties of exponents \\ = (1)^{34} \cdot i^{2} & i^{4} = 1 \\ = 1 \cdot - 1 & i^{2} = - 1 \\ = - 1 \end{aligned}

i^{138} = - 1

Now you might ask why we chose to write

i^{138}

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 $i^{- 1}$ ‍?

One easy way to approach this problem is to use and extend upon the pattern.

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

However, instead of extending the pattern to the right (for exponents

> 1

), we can think about extending the pattern to the left (for exponents

< 1

). In fact, since we want to know

i^{- 1}

, we can just extend the sequence two places to the left.

	$i^{- 1}$ ‍	$i^{0}$ ‍	$i^{1}$ ‍	$i^{2}$ ‍	$i^{3}$ ‍	$i^{4}$ ‍	$i^{5}$ ‍	$i^{6}$ ‍	$i^{7}$ ‍	$i^{8}$ ‍
			$i$ ‍	$- 1$ ‍	$- i$ ‍	$1$ ‍	$i$ ‍	$- 1$ ‍	$- i$ ‍	$1$ ‍

Filling in the blanks, we have:

$i^{- 1}$ ‍	$i^{0}$ ‍	$i^{1}$ ‍	$i^{2}$ ‍	$i^{3}$ ‍	$i^{4}$ ‍	$i^{5}$ ‍	$i^{6}$ ‍	$i^{7}$ ‍	$i^{8}$ ‍
$- i$ ‍	$1$ ‍	$i$ ‍	$- 1$ ‍	$- i$ ‍	$1$ ‍	$i$ ‍	$- 1$ ‍	$- i$ ‍	$1$ ‍

i^{- 1} = - i

We can also verify this algebraically by using properties of exponents as shown below:

\begin{aligned} i^{3} & = i^{4} \cdot i^{- 1} \\ - i & = 1 \cdot i^{- 1} & Since i^{3} = - i and i^{4} = 1 \\ - i & = i^{- 1} \end{aligned}

Want to join the conversation?

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satyanash
Posted 8 years ago. Direct link to satyanash's post “How is `i^3 = -i` ? My un...”
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
Button navigates to signup pageComment on satyanash's post “How is `i^3 = -i` ? My un...”
(52 votes)
Answer
- andrewp18
  Posted 8 years ago. Direct link to andrewp18's post “On step 3 you did: √𝑎 • ...”
  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.
  Comment on andrewp18's post “On step 3 you did: √𝑎 • ...”
  (150 votes)
dylan.forr99
Posted 10 months ago. Direct link to dylan.forr99's post “On hour 9 of study. send ...”
On hour 9 of study. send help. xD
Button navigates to signup pageComment on dylan.forr99's post “On hour 9 of study. send ...”
(39 votes)
Answer
- Vestige
  Posted 10 months ago. Direct link to Vestige's post “sending help your way _h...”
  sending help your way
  help sent
  Comment on Vestige's post “sending help your way _h...”
  (15 votes)
dollyrauh
Posted 8 years ago. Direct link to dollyrauh's post “Hi, could someone maybe h...”
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!!
Button navigates to signup pageComment on dollyrauh's post “Hi, could someone maybe h...”
(14 votes)
Answer
- Dominik.Ehlert
  Posted 8 years ago. Direct link to Dominik.Ehlert's post “It seems to me your probl...”
  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
  Comment on Dominik.Ehlert's post “It seems to me your probl...”
  (30 votes)
Jonathan Huang
Posted 7 years ago. Direct link to Jonathan Huang's post “Where do you use this in ...”
Where do you use this in the real world?
Button navigates to signup pageButton navigates to signup page
(14 votes)
Answer
- 🎓 Arnaud Jasperse
  Posted 7 years ago. Direct link to 🎓 Arnaud Jasperse's post “Physics uses it. Electric...”
  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).
  Button navigates to signup page
  (28 votes)
joey mizrahi
Posted 7 years ago. Direct link to joey mizrahi's post “how is i^-1 = -i doesnt i...”
how is i^-1 = -i
doesnt it equal 1/i
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Jacky Xie
  Posted 7 years ago. Direct link to Jacky Xie's post “Yes, i^-1 = 1/i. Notice t...”
  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.
  Button navigates to signup page
  (17 votes)
Rob
Posted 8 years ago. Direct link to Rob's post “will i^0=1?”
will i^0=1?
Button navigates to signup pageComment on Rob's post “will i^0=1?”
(9 votes)
Answer
- Abhishek Kumar
  Posted 5 years ago. Direct link to Abhishek Kumar's post “Hey, the explanation to y...”
  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.
  Comment on Abhishek Kumar's post “Hey, the explanation to y...”
  (6 votes)
Peu255
Posted a year ago. Direct link to Peu255's post “where are the powers of t...”
where are the powers of the imaginary unit? I've seen no powers. Does it throw fireballs?
Button navigates to signup pageComment on Peu255's post “where are the powers of t...”
(11 votes)
Answer
umairansari16
Posted 7 years ago. Direct link to umairansari16's post “what Will be i to the pow...”
what Will be i to the power i 899999
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- kubleeka
  Posted 7 years ago. Direct link to kubleeka's post “i^899999 (i^900000)/i 1/i...”
  i^899999
  (i^900000)/i
  1/i
  i/(i^2)
  i/(-1)
  -i
  Comment on kubleeka's post “i^899999 (i^900000)/i 1/i...”
  (12 votes)
O Madhu 💞
Posted 9 months ago. Direct link to O Madhu 💞's post “I understand how to solve...”
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.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Tanner P
  Posted 9 months ago. Direct link to Tanner P's post “I’ll walk you through an ...”
  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
  Comment on Tanner P's post “I’ll walk you through an ...”
  (18 votes)
owengeorge
Posted 10 months ago. Direct link to owengeorge's post “Is there any reason to us...”
Is there any reason to use multiple of 4 over multiple of 1,2 etc or is just arbitrary?
Button navigates to signup pageComment on owengeorge's post “Is there any reason to us...”
(4 votes)
Answer
- Kim Seidel
  Posted 10 months ago. Direct link to Kim Seidel's post “It must be 4. It is not a...”
  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.
  Comment on Kim Seidel's post “It must be 4. It is not a...”
  (10 votes)

Algebra 2