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### 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

The imaginary unit i is defined such that i²=-1. So what's i³? i³=i²⋅i=-i. What's i⁴? i⁴=i²⋅i²=(-1)²=1. What's i⁵? i⁵=i⁴⋅i=1⋅i=i. Discover how the powers of 'i' cycle through values, making it possible to calculate high exponents of 'i' easily. Created by Sal Khan.

## Want to join the conversation?

- What would happen if you rose i to a negative power, say -3?(255 votes)
- Negative exponents mean you take 1 over the value. So, for -3 you would get 1/i^3 = 1 / -i = -1/i.(291 votes)

- i've been doodling around with this and i found that: i^n = i^(n mod 4) ...

suppose you have i^549,

above i state that i^549 = i^(549 mod 4),

so we look at the remainder of 549 divided by 4 and we get 1 (549 mod 4 = 1),

=> i^549 = i^1 = i

hope it's helpful if i'm right, if not, i hope someone corrects me quick so i don't mislead somebody with my doodles.(88 votes)- Yes that's correct and very helpful :-)

I would also like to show and share the same process in detail...

To determine what happens to an imaginary number such as i when raised to a certain power.

Just follow these steps and you will be able to solve for the value of an imaginary number i raised to any power.**The following steps are as indicated below****1st**you divide the power of i by 4**2nd**you replace its power by the remainder of the original power of i**3rd**your answer will now be determined by the new power of your i [which is the remainder of the original power]

For instance:

if there is no remainder then the new power of i is zero so

i^0=1 Because (any number)^0=1

if the remainder is 1 then

i^1=i Since (any number)^1=itself

if the remainder is 2 then

i^2=-1

and finally, last but not least (and by the way this is as big as you're remainder will ever get), if the remainder is 3

i^3=-i Since (i^2)*i^1=-1*i=-i

For example: you have i^333 and you want to its value.

1st divide 333 by 4 and you get 83 1/4.

2nd you replace its power by the remainder which is 1, i^1

3rd its new power will determine its value

since the remainder is 1

i^1=i Since (any number)^1=itself

Just follow these steps and you will be able to solve for the value of an imaginary number i raised to any power.(73 votes)

- What if you have to multiply or divide i, like i^789*i^354 or i^766/i^32?(15 votes)
- You would use the rules of multiplying and dividing exponents. For your i^766/i^32 would get i^766-32 = i^734. Similarly with: i^789*i^354 = i^789+354 = i^1,143. You would then evaluate it.(23 votes)

- Sal mentions at the very beginning of the video that there is a cycle here: 1, I, -1, -I; We see a four stage cycle, in fact. Is there and relationship or connection say with the trigonometric functions (i.e. sin & cos) where we also see similar four stage cycles of 0, 1, and -1?(22 votes)
- Actually, yes. There is a pretty tight connection between complex numbers and trigonometry -- look up "polar form" of complex numbers under "complex plane" in this section.

The polar form for i is the quantity 1 * (cos (pi/2) + i*sin (pi/2)). Raising this to the nth power, according to DeMoivre's Theorem gives you the quantity 1 * (cos(n*pi/2) + i * sin(n*pi/2)). (Would love to explain, but is best if you visit those topics first.)

Further connections when you get to Taylor series in Calculus.(15 votes)

- I want to cry because of this(22 votes)
- Yeah, me too(3 votes)

- How would you explain what i^i would be?(16 votes)
- This answer is not correct. You have to use Euler's formula. i^i = {e^i(2kpi+pi/2)}^i = e^i^2(2kpi+pi/2) = e^-(2kpi+pi/2) where k is an element of the set of integers. The principle value for k=0 = e^-pi/2 = 0.207879576350761908546955465465465(7 votes)

- With the example when i was raised to the 100th power, how did you you know to take out a 4 out of the power? I understand this gets you the result of 1, which makes things easier. However, 100 is divisible by 5. This would return 20i. If you took out 2 from 100 you would get -1. On a test, how would one know what to use?(8 votes)
- But if use (i^5)^20 and i^5 is equal to i and so i^20, note that it is not multiplied by 20 it is exponent of i. Then again i^20 is (i^4)^5=1^5=1 only...(3 votes)

- Then, what is i to the power of x if x is a decimal or irrational number like 0.6?(5 votes)
- Well, can't we just divide 99 by 4 and get 24.75? Then doesn't it mean that it cycled 24 times and is now on its 3rd value? I mean in this it would be -1. But in this tutorial it is -i.

Is the technique I just described wrong?(4 votes)- You're on the right track. But, you are getting the wrong final result. There is a repeating cycle:

i^1 = i

i^2 = -1

i^3 = -i

i^4=+1

So, for 3, the result is -i, not -1

Hope this helps.(11 votes)

- what happens when we put 1 rise to the power i ?(2 votes)
- 1^i is just 1.

In fact, 1 raised to just about anything is just 1.

According to the geniuses at Wolfram, even "1^banana" is still 1. Go figure. :)(14 votes)

## Video transcript

Now that we've seen
that as we take i to higher and higher powers,
it cycles between 1, i, negative 1, negative i, then
back to 1, i, negative 1, and negative i. I want to see if we can
tackle some, I guess you could call them,
trickier problems. And you might see these surface. And they're also
kind of fun to do to realize that you can use
the fact that the powers of i cycle through these values. You can use this to really,
on a back of an envelope, take arbitrarily
high powers of i. So let's try, just
for fun, let's see what i to the
100th power is. And the realization here is
that 100 is a multiple of 4. So you could say that this
is the same thing as i to the 4 times 25th power. And this is the same thing, just
from our exponent properties, as i to the fourth power
raised to the 25th power. If you have something
raised to an exponent, and then that is
raised to an exponent, that's the same thing as
multiplying the two exponents. And we know that
i to the fourth, that's pretty straightforward. i to the fourth is just 1. i to the fourth is
1, so this is 1. So this is equal to
1 to the 25th power, which is just equal to 1. So once again, we use this
kind of cycling ability of i when you take its
powers to figure out a very high exponent of i. Now let's say we try something
a little bit stranger. Let's try i to the 501st power. Now in this situation, 501,
it's not a multiple of 4. So you can't just
do that that simply. But what you could do,
is you could write this as a product of two
numbers, one that is i to a multiple
of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as
i to the 500th power times i to the first power. Right? You have the same base. When you multiply,
you can add exponents. So this would be i
to the 501st power. And we know that this
is the same thing as-- i to the 500th power
is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right
over here. i to the 500th is the same thing as i to the
fourth to the 125th power. And then that times
i to the first power. Well, i to the fourth is 1. 1 to the 125th power
is just going to be 1. This whole thing is 1. And so we are just left
with i to the first. So this is going
to be equal to i. So it seems like a really
daunting problem, something that you would have
to sit and do all day, but you can use this cycling
to realize look, i to the 500th is just going to be 1. And so i to the 501th is just
going to be i times that. So i to any multiple of 4--
let me write this generally. So if you have i to any multiple
of 4, so this right over here is-- well, we'll just restrict k
to be non-negative right now. k is greater than or equal to 0. So if we have i to any
multiple of 4, right over here, we are going to get 1, because
this is the same thing as i to the fourth power
to the k-th power. And that is the same thing
as 1 to the k-th power, which is clearly equal to 1. And if we have
anything else-- if we have i to the 4k plus 1 power,
i to the 4k plus 2 power, we can then just do this
technique right over here. So let's try that with a
few more problems, just to make it clear that
you can do really, really arbitrarily crazy things. So let's take i to
the 7,321st power. Now, we just have
to figure out this is going to be some multiple
of 4 plus something else. So to do that, well, you could
just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have
that 1 left over. And so this is going to
be i to the 7,320 times i to the first power. This is a multiple of 4-- this
right here is a multiple of 4-- and I know that because
any 1,000 is multiple of 4, any 100 is a multiple of 4,
and then 20 is a multiple of 4. And so this right over
here will simplify to 1. Sorry, that's not i
to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over
here is going to simplify to 1, and we're just going
to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try
something interesting. i to the 99th power. So once again, what's
the highest multiple of 4 that is less than 99? It is 96. So this is the same thing
as i to the 96th power times i to the third power, right? If you multiply these, same
base, add the exponent, you would get i
to the 99th power. i to the 96th power, since
this is a multiple of 4, this is i to the fourth, and
then that to the 16th power. So that's just 1 to the
16th, so this is just 1. And then you're just left
with i to the third power. And you could either remember
that i to the third power is equal to-- you
can just remember that it's equal to negative i. Or if you forget that,
you could just say, look, this is the same thing
as i squared times i. This is equal to
i squared times i. i squared, by definition,
is equal to negative 1. So you have negative 1 times
i is equal to negative i. Let me do one more
just for the fun of it. Let's take i to the 38th power. Well, once again,
this is equal to i to the 36th times i squared. I'm doing i to the 36th
power, since that's the largest multiple
of 4 that goes into 38. What's left over is this 2. This simplifies
to 1, and I'm just left with i squared, which
is equal to negative 1.