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Integrated math 2

Course: Integrated math 2>Unit 5

Lesson 1: The imaginary unit i

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.

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• What would happen if you rose i to a negative power, say -3?
(248 votes)
• Negative exponents mean you take 1 over the value. So, for -3 you would get 1/i^3 = 1 / -i = -1/i.
(283 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.
(89 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.
(70 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)
• 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.
(22 votes)
• I want to cry because of this
(23 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
(6 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.
(10 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. :)
(12 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.