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Introduction to Imaginary Numbers - Concept
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An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i^2 = -1. **Imaginary numbers** are applied to square roots of negative numbers, allowing them to be simplified in terms of i. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Beware that in some cases the letter j is used instead of i for the imaginary number.

Intro to imaginary numbers. Now imaginary numbers contrary to their name actually do exist, sort of a weird concept that we're going to be getting into.

So to start that out, let's do a little bit of a review on radicals. So say I say something like square root of 8, and I say simplify that. Your logic is like okay. What goes into 8. We have square root of 4 times 2. Square root of 4 is a perfect square so we can take that out giving us 2 root 2. Where imaginary numbers come in is when we are taking the square root of a negative number. Okay? And up until now what we've said is you can't take the square root of a negative number which is sort of a lie, I apologise. But in order to do that we have to use these imaginary numbers, okay?

So take a look at the square root of -9. Okay. We've been saying this isn't real which is true, okay? But just like we did up here, we can split this up into a real component which is the 9 part we know and the -1 part which is the part that sort of throws us off. Okay?

So what we're going to say is this is the same thing as the square root of 9 times -1. We can split up square roots and then write it in up here like you know how to do it. So this is the square root of 9 times the square root of -1. Square root of 9 is 3. So what we're really left with here is 3 times the square root of -1. This is where imaginary numbers come into play. And square root of -1 is really the basis of all imaginary numbers we deal with. And going to give you a little bit of definition. Okay?

What we do is we call the square root of -1 the letter i. Lower case case of i. So going back to the problem we have over here we can replace that in, and this ends up being 3i, i standing for the imaginary number, square root of -1. Okay. In addition there is one other part of the definition of i, okay? And that is i squared.

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