Tuesday, August 6, 2013

IMAGINARY NUMBERS, TOOLS AND TRICKS


I tell my students that mathematicians don't like situations in which they can't find an answer, so when the quadratic equation so frequently comes up with a negative discriminant, it drives them crazy.  ("You can't take the square root of a negative number!!," emphatically warns the arithmetic teacher.)

As a mathematical cure for the insanity, someone introduced the "imaginary" number.  It's like your little brother's imaginary friend; you can see it plainly in your head, use it for all kinds of activities and excuses, but nobody else can actually put their finger on it.  It's purpose is to take the negativity out of a square root and make the problem solvable.

√ (-49) = √(-1)√(49)
So make the square root of negative one equal 'i' and the problem disappears.

√ (-49) = 7i

And with just that little piece of information, I'm ready for the quiz.
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That was all very easy, so let's look at some "tricks."

If   i = √(-1)
then i^2 = -1
and i^3 = -1 √(-1) = -1i = -i
and i^4 = (-1)(-1) = 1

So any i raised to a power evenly divisible by 4 equals 1.  This reminds me of the trick about 0 as an exponent:  (almost) Any base raised to the zero power equals 1.  I love seeing that on tests in problems that look like...

     (43x^90 + 103x^80y^15 - 74y^101) ^ 0

                    = 1

With i's, breaking it into parts equal to i^4 is like taking out '1' many, many times, and 1 times 1 times 1....still equals one.

To find i to any other power, just attach the remainder to i.
     i^81 = i^1         81÷ 4 = 20 (who cares?) with a remainder of 1
     i^ 27 = i^3        27 ÷ 4 = 6 (useless information) with a remainder of 3
     i^98 = i^2= -1   98 ÷ 4 = 24 (yawn) with a remainder of 2
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How easy was THAT?!?  Give it to me on a quiz and I'm sure to get that point!!

Multiplying conjugates that contain imaginary numbers is also a no brainer.  The middle terms cancel each other out and you're left with squaring i to make it a rational number.

     (x - 4i)(x + 4i) = x^2 – 4ix + 4ix - 16i^2
          Let i^2 = –1
     (x – 4i)(x+ 4i) = x^2 + 16  (Catch the next blog on "breaking rules.")

This little trick comes in handy when you are confronted with a complex number in the denominator:


And BEHOLD...the denominator turns into an integer.

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There's one more trick that is handy for the tool kit of aspiring Math nerds and others who want to be a pain in the side of any math teacher.  But that one will be posted separately so you have to look for it and prove that you're really committed to creating waves, because it is BREAKING THE RULES!!! (sort of)

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