Imaginary (Complex) NumbersDate: 02/26/98 at 10:00:18 From: Chris Subject: Imaginary (complex) numbers What are imaginary numbers? I have to do a report on them and can't find anything understandable anywhere else. Thank you for your time. Date: 02/26/98 at 15:25:10 From: Doctor Rob Subject: Re: Imaginary (complex) numbersz You probably are aware that for any real number a, a^2 >= 0. If one is faced with an equation like x^2 = -9, one can immediately conclude that there is no real solution. Imaginary numbers were invented to provide solutions of equations like this. The simplest imaginary number is called i. It satisfies the property that i^2 = -1. Using it, the solutions of x^2 = -1 can be found: x = i is one, and x = -i is another, if we define -i properly. Since i cannot be a real number, it must be something else. It is called an "imaginary" number. Of course, it is no more imaginary than 7, 0, -4, 3/5, sqrt(2), pi, or any other number. They are all figments of our imagination! In order to be a "number," i must satisfy the axioms of arithmetic. It must have an additive inverse, or negative, or opposite, -i, so that i + (-i) = 0. It must have a multiplicative inverse, or reciprocal, which it does: in fact -i = 1/i. (You can prove this by showing that i*(-i) = 1.) Thus -i is both the negative and the reciprocal of i. Multiplication by a real number must be defined, so a*i must make sense, for any real number a. In particular, (-1)*i = -i. All the other axioms of arithmetic must also be true for the set containing the real numbers and i and whatever else you can get by doing arithmetic with them. Now you can solve x^2 = -9 = 9*(-1) = 3^2*i^2 = (3*i)^2. The two solutions are x = 3*i and x = -3*i, imaginary numbers! The same trick works for x^2 = -a, where a > 0 is a real number. x^2 = -a = a*i^2, so x = sqrt(a)*i and x = -sqrt(a)*i are the two solutions, imaginary numbers. The system of numbers you end up with by looking at the set containing the real numbers and i and whatever else you can get by doing arithmetic with them has a name, of course. It is called the complex numbers. It turns out that every complex number can be written in the form a + b*i, for some real numbers a and b. This means that every complex number is the sum of a real number and an imaginary number. The rules for doing arithmetic with complex numbers are all the ordinary ones, because we insisted that that be so when we built up the set. The only difficulty in seeing that the form a + b*i includes all complex numbers is in multiplication and division. For multiplication, do the ordinary multiplication of two complex numbers of this form using FOIL, then use the fact that i^2 = -1, and combine terms containing i and ones not containing i. For example, (3 - 2*i)*(-1 + i) = -3 + 3*i + 2*i - 2*i^2, = -3 + 3*i + 2*i + 2, = -1 + 5*i. The result is back in the same form, a real number -1 plus an imaginary number 5*i. For division, we have to know one trick: (a + b*i)*(a - b*i) = a^2 + b^2. Multiplying these two complex numbers gives a real number answer! If you find this hard to believe, multiply it out using FOIL and i^2 = -1, and see what you get. Now we can do division. To divide by a + b*i, instead multiply by its reciprocal, which is (a - b*i)/(a^2+b^2) = a/(a^2+b^2) + [-b/(a^2+b^2)]*i, and use the rules given above to reduce it back to the form of a real number plus an imaginary number. Of course, we cannot do this if a^2 + b^2 = 0, but that only happens if a = b = 0, so the divisor is 0 + 0*i = 0. Division by zero is forbidden, but we are used to that! As an example, since (-1 + i)*(-1 - i) = (-1)^2 + 1^2 = 2, and so the reciprocal of -1 + i is (-1 - i)/2 = (-1/2) + (-1/2)*i, we get (3 - 2*i)/(-1 + i) = (3 - 2*i)*(-1 - i)/2, = (-3 - 3*i + 2*i + 2*i^2)/2, = (-5 - i)/2, = -5/2 - (1/2)*i. Now you can do any arithmetic with complex numbers, and in particular, with imaginary numbers. You can also take square roots of any real number, whether positive, zero, or negative. -Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/