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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/   
    
Associated Topics:
High School Imaginary/Complex Numbers

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