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What is an Imaginary Number?

What is an imaginary number?

Summary

  1. 'b' is a variable representing a non-zero real number
  2. 'i' is NOT a variable -- it is the imaginary unit and is equal to the square root of -1
  3. The equal sign with a line through it means 'not equal'
  4. The Product Property of Square Roots allows you to break the square root into separate square roots
  5. The square root of -4 is equal to the imaginary number 2i
  6. i2 is equal to -1, a real number!
  7. 'r' is just a variable representing any non-zero real number

Notes

    1. A quadratic equation has the form ax2+bx+c = 0
    2. 'a' is the coefficient in front of x2
    3. Since there is no coefficient in front of x2, we have an invisible coefficient of 1
    4. 'b' is the coefficient in front of 'x', so 'b' is -4
    5. 'c' is the constant term, so 'c' is 5
    6. Plug 1, -4, and 5 into the Quadratic Formula for 'a', 'b', and 'c'
    7. We end up with a negative under the square root!
    1. We get the square root of -4 as part of our answer when we simplify
    2. There are no real numbers we can square to get -4, so that means we have no real solutions
    3. You might have been told that you can't take the square root of a negative number
    4. You actually CAN take the square root of a negative number, but you'll get what's called an 'imaginary number'
    1. Complex numbers can be written in the form a + bi
    2. When 'a' is 0, then we have a complex number of the form 'bi', which we just call an imaginary number
    3. 'a' and 'b' are both real numbers, and 'i' is the imaginary unit
    4. 'i', the imaginary unit, is equal to the square root of -1
    1. Here 'b' represents a non-zero real number
    2. 'i' is the imaginary unit equal to the square root of -1
    1. 'a' and 'b' represent non-zero real numbers
    2. 'i' is the imaginary unit equal to the square root of -1
    3. 'b' is a real number, but when it's multiplied by 'i' it becomes part of the imaginary part
    1. 5 is a real number, but when it's multiplied by 'i' the whole thing becomes an imaginary number
    2. Remember, 'i' is the imaginary unit and is equal to the square root of -1
    3. Imaginary numbers can be negative too, but they're still imaginary!
    4. We can also multiply 'i' times a square root to get an imaginary number -- square roots are real numbers, so it still works!
    5. Notice that the 'i' comes before the square root so that it reads more easily -- it's still the same thing as 'bi'
    1. Recall from our first example that we got hung up from having the square root of a negative number in our solution
    2. We can use imaginary numbers to help us deal with that!
    3. -4 is the same thing as -1 times positive 4
    4. Since we have two things being multiplied under one square root, the Product Property says we can rewrite this as two roots being multiplied
    1. Recall that 'i' is equal to the square root of -1, so we can substitute the square root of -1 with 'i'
    2. 22 equals 4, so the square root of 4 is 2
    3. Since imaginary numbers have the form 'bi', it makes sense to rewrite i•2 as 2i
    1. Remember that we started with the square root of -4 and ended up with 2i
    2. Since 2i and the square root of -4 are the same thing, we can substitute 2i for √(-4)
    3. We can cancel a 2 in the numerator and denominator to get 2±i, which means our solutions are 2+i and 2-i
    1. Recall that we tried to solve this equation but had to deal with the square root of -4 in our solution
    2. Our solution was 2±i, so we can rewrite this as two separate solutions
    1. 2i is an imaginary number because it has the form 'bi'
    2. Remember, 'i' is the imaginary unit and is equal to the square root of -1
    3. Even though 'i' is NOT a variable, we can multiply it as if it were
    4. So i•i gives us i2
    5. Squaring √(-1) cancels out the square root, leaving us with just -1
    6. So i2 is equal to -1
    7. Since i2 equals -1, we can subtitute -1 for i2
    8. We squared an imaginary number and ended up with a real number -- this could be handy!
    1. 'r' could be any non-zero number here
    2. Even though the 'i' is in front of the √(r), it's still an imaginary number
    3. We just put the 'i' out front so it doesn't look like it's under the square root
    4. Again, we get i2 when we multiply i•i
    5. Multiplying √(r)•√(r) is the same as squaring √(r), so again the square root cancels out and we're left with just 'r'
    6. And again we can substitute -1 for i2, since we know they're the same thing
    1. Remember, 'i' is the imaginary unit and is equal to the square root of -1