A Primer on Complex ArithmeticDate: 10/29/2002 at 18:31:36 From: Lindsey Subject: Complex Numbers Hello! I really don't understand how to use complex numbers. How do I do ones like this: (4-10i)(4+10i)? Ones like this are even worse: w = 3-4i z = 3+4i Please help me! Lindsey Date: 10/29/2002 at 19:28:28 From: Doctor Ian Subject: Re: Complex Numbers Hi Lindsey, Here's a quick primer on complex arithmetic. For purposes of adding and multiplying, you can think of complex numbers as binomials, like (x - 2) or (x + 5). To add (2x - 2) and (3x + 5), you just add the coefficients of like terms: (2x - 2) + (3x + 5) = x(2 + 3) + (-2 + 5) = 5x + 3 (Similarly for subtraction.) To multiply (x - 2) and (x + 5), you just apply the distributive property a couple of times: (x - 2)(x + 5) = x(x + 5) - 2(x + 5) = x^2 + 5x - 2x - 10 = x^2 + 3x - 10 Same thing with complex numbers: (2 + 3i)(4 - 5i) = 2(4 - 5i) + 3i(4 - 5i) = 8 - 10i + 12i - 15i^2 = 8 - 2i - 15i^2 What makes this nicer, though, is that the i^2 term disappears, because i^2 is just equal to -1, = 8 - 2i + 15 = 23 - 2i giving you a new complex number. Do you remember this pattern? (x + a)(x - a) = x^2 - a^2 ? It works for complex conjugates, too: (3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - (4^2 i^2) = 9 - (-16) = 25 So when you multiply a complex number by its conjugate, you get a real number. Division looks scarier, but it's kind of like factoring. Suppose you want to find 26 - 7i ------- 3 + 4i When you want to factor something like x^2 - 2x -6, you start by assuming that it will look like (x + a)(x + b), and then you figure out what a and b are, right? Same thing here. You assume that you're going to end up with (a + bi), and then you use the definition of division: 26 - 7i ------- = a + bi 3 + 4i 26 - 7i = (a + bi)(3 + 4i) = 3a + 4ai + 3bi + 4bi^2 = (3a - 4b) + (4a + 3b)i ^ ^ | | So this So this has to has to equal 26 equal -7 Does this look familiar yet? That gives you two equations in two unknowns, 3a - 4b = 26 4a + 3b = -7 which you can solve by elimination or substitution. I'll use elimination: 9a - 12b = 78 (Multiply the 1st eq. by 3) 16a + 12b = -28 (Multiply the 2nd eq. by 4) ----------------- 25a = 50 So a = 2, and 3(2) - 4b = 26 6 - 26 = 4b -20 = 4b So b = -5, which means that 26 - 7i ------- = (2 - 5i) 3 + 4i This isn't the prettiest way in the world to do arithmetic. On the other hand, I'd rather do this than long division any day. The most important thing is not to panic. Keep in mind that if you add, subtract, multiply, or divide two complex numbers, you're going to end up with another complex number; and remember that i^2 = -1, so the exponents don't keep growing the way they do with polynomials. Let me know if any part of this didn't make sense, or if you have other questions. - Doctor Ian, 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/