The Imaginary Number "i"
Purplemath
Up until now, you've been told that you can't take the square root of a negative number. That's because you had no numbers which were negative after you'd squared them — so you couldn't "go backwards" and return to them by taking the square root.
Before now, every number was positive after you squared it. So you couldn't very well take the square root of a negative and expect to come up with anything sensible.
Now, however, you are told that you can take the square root of a negative number, but doing so requires using a new number. This new number was invented ("discovered"?) around the time of the Reformation. At that time, nobody believed that any "real world" use would be found for this new number, other than easing the computations involved in solving certain equations, so the new number was viewed as being a pretend number invented for convenience's sake. But—
When you think about it, aren't all numbers inventions? It's not like numbers grow on trees; they live in our heads. We made them all up. So why not invent a new one, as long as it works okay with whatever else we already have?
So they invented a new one.
This new number was called "i", standing for "imaginary", because "everybody knew" that i wasn't "real". (This, by the way, was why you couldn't take the square root of a negative number before: you only had "real" numbers to work with; that is, numbers without the "i" in them.)
What is the imaginary number?
The imaginary is defined to be:
When we square this new number, we get
![i^2 = (sqrt[-1])^2 = -1](complex/complx02.gif){kind=small}
Clearly, the imaginary number is a new thing, and brings new flexibility: we can now do something that before we couldn't; namely, we can take the square root of −1.
Now, based on past experience, you might think you can do this:
![i^2 = (sqrt[-1])^2 = WRONG! sqrt([-1]^2) = WRONG! sqrt[1] = 1 WRONG!](complex/complx03.gif){kind=small}
But this doesn't make sense in the context of the mathematics that we already had — and remember, this new number is only helpful as long as it works with what we'd already had. So how does this not make sense?
One reason this won't work is that we already have two numbers that square to 1; namely −1 and +1. What is the point of i, if it just duplicates stuff we already have? We want it to do new stuff.
Secondly, by definition, i squares to −1. So it is simply not reasonable to think that i would also square to +1. Since −1 can never equal +1, the "equation" above cannot be true.
But where did that equation go wrong? We've done this "move the power inside the radical" thing before, and it was okay. Where did the math go wrong this time?
It went wrong in the middle line, where the squaring was taken from outside the square root to inside. And this points out the limitation that comes with the imaginary number.
What do we gain / lose with the imaginary?
When dealing with imaginaries, we gain something (the ability to deal with negative numbers inside square roots), but we also lose something (being the flexible and convenient rule we used to have when dealing with the squaring of square roots). In particular, when simplifying with the imaginary — or, which is the same thing, when simplifying with the square root of −1 — WE MUST ALWAYS DO THE i-PART FIRST!
What is an example of simplifying with the imaginary number?
- Simplify
The number inside the square root is negative. To deal with this, I have to break apart the nine and the "minus", so my first step in my simplification will be to factor. Then I can move the "minus" into its own square root, and simplify by using the imaginary. My work looks like this:
Notation advisory: In going from the second line to the third, the conversion is "
", not "
". The i is outside and after the square-root symbol; it does not hop back in and under with the 9.
Don't make the mistake of closing up that space between the two square roots, so that the imaginary accidentally somehow climbs back inside of a square root. The imaginary is outside of that square root of nine, and it needs to stay there!
Let's do some more simplifying.
- Simplify .
I will split the "minus" from the 25, split the one square root into two, convert , simplify the other square root, and put things back together at the end for my answer.
- Simplify
This square root looks similar to the previous one, but there is an important difference; namely, 25 = 52 in the previous exercise was a perfect square, but the 18 = (32)×(2) is not. How to deal with this?
I'll do the simplification in the same way as I always have for the 18, and I'll break out the −1 and deal with it separately.
- Simplify
This square root has a "minus" sign in front of it. But, while I can take a "minus;" through parentheses, I can do no such thing with a radical. The "minus" on the outside of the radical is different from the "minus" on the inside, and I'll be dealing with them separately.
I note that the number inside the square root is not a perfect square, nor does it contain any. There is nothing to simplify with the 6. But I can work with that "minus" on the inside. So I'll split that apart to get the imaginary outside the radical, and the leading "minus" on the outside will be carried along for the ride.
Honestly, the only "simplifying" happening here was in converting to the imagainary form.
How do you add and subtract with i?
In your addition and subtraction computations with imaginaries, you will deal with i just as you would with x.
- Simplify 2i + 3i
If they'd given me 2x + 3x, I'd have just added the 2 and the 3, and carried the x along, for an answer of 5x.
Intead, they've given me an expression with the letter i. I'll deal with this in the exact same way.
2i + 3i = (2 + 3)i = 5i
- Simplify 16i − 5i
This one works just like the previous one.
16i − 5i = (16 − 5)i = 11i
How do you multiply with the imaginary?
To multiply expressions containing the imaginary i, multiply in the same way as you would for expressions containing a variable, like x. The only difference here is that x2 is just x2, but i2 is −1, so there is more simplification that can be done.
- Multiply and simplify
(3i)(4i)
To do this simplification, I will move the factors around, so that the numerical portions and the imaginaries are grouped together. Any squares of i will be converted to −1 and then multiplied into the numerical portion.
(3 i) (4 i) = (3 · 4)(i · i)
= (12) ( i2 )
= (12)(−1)
= −12
- Multiply and simplify (i)(2i)(−3i)
This multiplication has three copies of i, but that's okay. Two of them will multiply to −1, and the last copy of i will be carried along.
( i )(2 i )(−3 i ) = (2 · −3)(i · i · i)
= (−6)( i2 · i )
=(−6)(−1 · i )
= (−6) (−i)
= 6i
In the last exercise above, we had an i3 as part of the expression. In the process of simplification, this i3 simplified as i3 = −i. This simplification occurred because i2 = −1. We can continue to larger powers of i:
i4 = i2 · i2 = (−1)(−1) = 1
And just like that, we're back to 1 again.
This pattern of powers, signs, 1's, and i's is a cycle, as you can see below:
i1 = i
i2 = −1
i3 = −i
i4 = 1
i5 = i4 × i = 1 × i1 = i
i6 = i4 × i2 = 1 × i2 = −1
i7 = i4 × i3 = 1 × i3 = −i
i8 = i4 × i4 = 1 × 1 = 1
In other words, to calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent. For example, a common trick question on tests is something along the lines of "Simplify i99", the idea being that you'll try to multiply i ninety-nine times and you'll run out of time, and the teachers will get a good giggle at your expense in the faculty lounge. Here's how the shortcut works:
i99 = i96+3 = i(4×24)+3 = i3 = −i
That is, i99 = i3, because you can just lop off the i96. (Ninety-six is a multiple of four, so i96 is just 1, which you can ignore.) In other words, you can divide the exponent by 4 (using long division), discard the answer, and use only the remainder. This will give you the part of the exponent that you care about. Here are a few more examples:
- Simplify i17.
The power is one more than a multiple of four: 17 = 16 + 1 = 4×4 + 1. I will use this to reduce the power to something more reasonable:
i17 = i16 + 1
= i4 · 4 + 1
= i1
= i
- Simplify i 120.
The exponent here is pretty big, but I can see right off that it's a multiple of four: 120 = 4×30. So dividing the power by four leaves no remainder; I've got thirty copies of i4 = 1. Simplification is a doddle:
i120 = i 4 · 30
= i 4 · 30 + 0
= i 0
= 1
- Simplify i 64,002.
Okay; this power is ridiculous. When I see something off-the-charts like this, I think about what tricks I might be able to apply to make my life easier. In this case, the trick with the powers will get me home.
i 64,002 = i 64,000 + 2
= i4 · 16,000 + 2
= i2
= −1
Until now, you've been working with "real" numbers. Now you've seen how to work with imaginarie numbers. What do we get if we put them together? We get something called "complex" numbers.
Complex numbers have two parts, a "real" part (being any "real" number of the sort that you're used to dealing with) and an "imaginary" part (being any number with an "i" in it).
The "standard" format for complex numbers is "a + bi"; that is, real-part first and i-part last.
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