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A Primer on Complex Arithmetic

Date: 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/ 
Associated Topics:
High School Imaginary/Complex Numbers

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