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## College Algebra

### Course: College Algebra>Unit 6

Lesson 1: The imaginary unit i

# Intro to the imaginary numbers

Sal introduces the imaginary unit i, which is defined by the equation i^2=-1. He then gets to know this special number better by thinking about its powers. Created by Sal Khan.

## Want to join the conversation?

• What happens when you put i to the power of i?
• If you've learned about Euler's formula, you'd know that e^(ix) = cos(x) + i*sin(x) (Khan has videos about this formula). Say we want to find a value for x which makes this equal to i, so want cos(x) = 0 and sin(x) = 1. x = pi/2 is a value which satisfies this. So e^(i*pi/2) = i.

So now that we have this seemingly more complicated way of writing i, we can substitute it for the base of our exponent and raise it to the power of i. So we get (e^(i*pi/2))^i.

Recall that (a^b)^c = a^(b*c), so we can apply the same idea here. (e^(i*pi/2))^i = e^(i*i*pi/2) = e^(i^2 * pi/2).

Now by definition i^2 = -1 so that cancels out and we get i^i = e^(-pi/2), which is approximately equal to about 0.207...

Interesting how an imaginary number raised to the power of an imaginary number results in a real number.
• How can i^2 be equal to a negative number if a square cannot equal a negative? Why is i useful? How come it was invented?
• You're question is the answer. There is no "real" number that you could find on the number line that when squared would equal a negative number, so mathematicians created (in their diabolical laboratory of weird and mystical things) a number that would fulfill those properties. In terms of usefulness, you'll see i pop up in quadratics that don't touch the x-axis and later if you do work in the complex plane. It's also in some advanced calculus and other topics, so don't think you can run away from it just because it's not real ;)!
• How could I apply Imaginary Numbers to the real world, what fields require knowledge of these numbers?
• Electrical engineering uses complex numbers to represent current, voltage, impedance since those quantities have both magnitudes and phase. Physics uses complex numbers for the same reasons.
• Whats an example of a real life problem related to Imaginary numbers? Something usefull as apposed to a crazy theory someone came up with because they didn't understand negative roots...
• complex numbers(numbers with i in them)are great for dealing with things that are periodic due to the properties mentioned in this vid.
For example, something oscillating back and forth on a spring can be thought of in terms of complex numbers. On the other hand, you can also describe it using sines and cosines... but sometimes it just works out simpler to do it the complex way (as the above poster mentioned, there are cases where the only way to get the answer is going through complex analysis, and arriving at a real answer).
• Is there any way to find a pattern in the power of i? For example, if a friend were to ask me to find the answer for i to the power of 542, is there a pattern in the power I can follow, or is the only way to count up by multiples of 4 (since the pattern is 1, i, -1, -i, repeat).
• Counting up by multiples of 4 can be achieved by dividing by 4.
The pattern is i, i^2=-1, i^3=-i, i^4=1
When you divide by 4, the remainder will always be either 0.25, 0.5, 0.75, or 0.
0.25 means i
0.5 means i^2=-1
0.75 means i^3=-i
and no remainder (0) means i^4 = 1
So, given i^542, divide 542 by 4, to get 135.5, so the answer is i^2=-1
Lets try i^333. 333/4= 83.25, so the answer is i.
and i^7, we have 7/4=1.75, so the answer is i^3=-i.
Hope that helps.
• At Sal says E. What is that number and can i have a link to it?
• e is a constant that comes up in math and science all the time. It is an irrational and transcendental number the first few digits of which are 2.71828...

e is officially defined as:
lim h→0 (1+h)^(1/h)
This same definition can also be expressed as:
lim h→ ∞ (1+1/h)^(h)
• At Sal says E. Now I know what it is @ Just Keith. But when and why do you use it? Is it like pi (3.1415926535897932)
• e is used in a variety of ways. One of the most common ways it is used is in the natural log. Natural logs are logarithms base e. They become more useful once you get to calculus, namely, the antiderivative of ln(x) is 1/x (which comes in handy a lot!).
I don't know what you mean by 'Is it like pi' but I think you mean "is it irrational" which it is. Some of its digits can be found here: http://apod.nasa.gov/htmltest/gifcity/e.2mil
By the way, also read Just Keith's answer (which is more in-depth than mine).
Hope this helps!
• @-58 Sal discusses:
i^4=(i)(i^3)=(i)(-i)=(-1)(i)(i)...
-How does the equation go from (i)(-i) to (-1)(i)(i)?
I tried to grasp the concept by studying how he defines (i^3) but it isn't quite clicking? If someone could further break this down for me that would be wonderful.
• To explain this, first you must know that i^2 is equal to -1 because i=√(-1) and √(-1)* √(-1) cancels out the √ and therefore gives you a -1. So back to the question, i^4 can be broken down into (i)(i^3) because if you multiply (i) with (i^3) since they have the same base it would be just be i^(1+3) which would give i^4. The (i)(i^3) can be broken down into (i)(-i) because remember i=√(-1), if you multiply √(-1) with √(-1) ( which we got -1 before) and multiply it once more with i, then you get a negative i (because -1*i= -i). Then, afterwards you can break down the -i to become (-1)(i) and you can add the part above which was i to give you an answer (-1)(i)(i). There are two i’s and as explained above i^2=-1. To find the final answer, by multiplying the -1 with the -1, you will get an answer of 1. Hope this helps!
• So for i to the power of 4 (i^4),

we're basically multiplying i^2 by i^2,

which is (-1)(-1),

which equals 1.

Right?
• You are absolutely right! Since i = sqrt(-1), i^2 is equal to -1. (-1)(-1) is indeed 1. Good job!
• Wouldn't something to the zero-th power be zero?
• This subject was confusing for me so I will offer another way of looking at it than kubleeka.

a^2 is a*a
a^3 is a*a*a
and this keeps going.

If you worked backwards though you might notice you divide by a to get to the previous 1.

so a^3 / a = a^(3-1) = a^2 or a*a*a/a = a*a
if we keep going it goes like this:
a^2 / a = a^(2-1) = a^1 or a*a/a = a
Then:
a/a = a^(1-1) = a^0 or a/a = 1

Then you could keep going to negative powers.