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## Algebra 2

### Course: Algebra 2>Unit 2

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. This pattern repeats itself all the time, so we can quickly tell what iⁿ for any positive integer n. Created by Sal Khan.

## Want to join the conversation?

• What would happen if you rose i to a negative power, say -3?
• Negative exponents mean you take 1 over the value. So, for -3 you would get 1/i^3 = 1 / -i = -1/i.
• 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.
• 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.
• What if you have to multiply or divide i, like i^789*i^354 or i^766/i^32?
• 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.
• 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?
• 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.
• How would you explain what i^i would be?
• 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
• Then, what is i to the power of x if x is a decimal or irrational number like 0.6?
• 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?
• 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...
• what happens when we put 1 rise to the power i ?
• 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. :)
• cant every power of i equal 1? for example lets say we have i^25 , we can write it as
i^(4)*(25/4) = (i^4)^25/4 = 1^(25/4) = 1?
• Great question! This type of situation shows that the exponent law (x^a)^b = x^(ab) does not always work for complex numbers x, if a and b are not both integers.

Have a blessed, wonderful day!