Multiplying and Dividing Complex Numbers
Date: 07/16/98 at 12:19:31 From: Michael Subject: Multiplying and dividing complex numbers Dear Dr. Math, How do you calculate (a+bi)*(c+di) and (a+bi)/(c+di)? Appreciate any information, Mic
Date: 07/24/98 at 10:29:05 From: Doctor Erra Subject: Re: Multiplying and dividing complex numbers Hello Michael, I'm assuming that i is sqrt[-1]. For the first question, I'll demonstrate the process with an example. Let us compute (2+3*i)*(5+7*i). We can expand this product: (2+3*i) * (5+7*i) = 2*5 + 2*7*i + 5*3*i +5*7*i*i = 10 +14*i +15*i +35*i^2 Now we use the fact that i^2 = i*i = -1, and with classical properties of algebraic expressions, we get: (2+3*i) * (5+7*i) = 10 + 29*i - 35 = -25 + 29*i The computation is exactly the same with (a+bi)*(c+di), and I'll let you complete this. The other expression is somewhat different, we have: (a+bi)/(c+di) Implicitly we want to find some real numbers x and y such that: (a+bi)/(c+di) = x+y*i I will give you the way, we first use the fact that: x/y = (x*z)/(y*z) (Equation 1) for all x, y, z such that the expression is "legal", which means that z cannot be 0. Complex numbers, such as a+b*i with a and b real, have a "conjugate" complex number, defined as a-b*i. One of the numerous properties that connect a complex number and its conjugate is: (a+b*i)(a-b*i) = a^2+b^2 To find x and y such that (a+b*i)/(c+d*i) = x+y*i, you can combine together the identity in Equation 1 and a good choice for z. Choose the conjugate of (c+d*i), which is (c-d*i) for z, so you have to compute: (a+b*i)/(c+d*i) = (a+b*i)/(c+d*i) * (c-d*i)/(c-d*i) = ... I leave the final computations to you. - Doctor Erra, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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