Simplifying Complex Numbers

Date: 05/23/2003 at 17:52:26
From: Meggen Stockstill
Subject: Simplifying imaginary numbers

When doing the problem: sqrt(-49) * sqrt(-16), if you simplify the 
radicals you get 7i * 4i = 28 i^ 2, which equals -28. However, if you 
use the Square Root of a Product Rule, you get sqrt(-49) x sqrt(-16) = 
sqrt(-1 * -1 * 49 * 16), which equals sqrt(49 x 16), giving you an 
answer of +28.

I know the answer should be -28, but this seems to contradict the 
Radical rule that states sqrt(ab) = sqrt(a) x sqrt(b). Can you explain 
why the Product Rule doesn't apply? I am an 8th grade algebra teacher 
and we couldn't explain this apparent contradiction in rules.  Since 
math is usually consistent, can you tell me if the Product Rule only 
holds for real numbers or what the explanation is? 

Date: 05/23/2003 at 18:16:52
From: Doctor Tom
Subject: Re: Simplifying imaginary numbers

Hi Meggan,

Unfortunately, the problem is a bit deep. It arises because every 
number (real or complex) other than zero has TWO square roots.

If we agree to stick with the reals, there's no problem: we agree to 
use the sqrt symbol to represent the POSITIVE root of positive reals, 
but we know that there's an additional root that is simply the 
negation of that number.  Also, sticking to the reals only, there IS 
NO square root of negative numbers.

Once you open the door to complex numbers, you've actually opened a 
mini-Pandora's box: non-zero numbers have two square roots, three cube 
roots, four fourth roots, and so on.  Worse, numbers have an infinite 
number of logarithms.

Although the imaginary number i "looks" positive, there is absolutely 
no reasonable way to split the complex numbers (or even i and -i) into 
"positive" or "negative" parts.  Say you decide that the square root 
of -1 is i.  Then the square root of -1 + .001i is quite close to i.

As you move continuously around on the plane (avoiding zero), if you 
take tiny steps, the answer you get will change by tiny amounts. But 
if you take a series of tiny steps from -1 in a path in the complex 
plane around zero, although you started with i as the square root, you 
arrive at -i after one loop.

What this means is that there is NO continuous way to assign square 
roots to complex numbers over the entire plane.

Basically what this means is that the rule: sqrt(AB) = sqrt(A) * 
sqrt(B) does not hold for complex numbers.  It might work for small 
sets of them, but it can't work for the entire complex plane.

If you looked at cube roots, the situation would be even worse.  For 
square roots, your "error" in applying that formula will simply amount 
to a possible factor of -1.  For cube roots, if you try to apply the 
formula cube_root(AB) = c_r(A) * c_r(B), depending on how you do it, 
you can get three different answers that differ by factors of 
(-1+sqrt(3)i)/2 or (-1 -sqrt(3)i)/2 - the two other cube roots of 1.  
For a similar attempt with fourth roots, the error factor might be -1, 
i, or -i, depending on the numbers you pick.  The higher the root, the 
worse the problem.

The problem can be "fixed" if you're willing to look at something 
called "Riemann surfaces" - weird topological spaces that are locally 
just like the complex plane but globally have strange connectiveness 
properties.  I don't know if you studied complex variables in 
university, but that's where you'd stumble across these things.

So, bottom line: the formula

  n-th_root(AB) = n-th_root(A) * n-th_root(B)

ONLY holds for non-negative real A and B, where you agree always to 
take the positve n-th root.

What makes it seductive is that it "almost" works for all the rest of 
the complex numbers, and it works for all of them if you're willing to 
allow a constant factor error which is always going to be one of the 
n-th roots of 1.

Sorry it's not simpler, but that's the way it is.  Unless the student 
is really good, just say that the restricted form of the rule is all 
that holds.

Good luck!

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/25/2003 at 17:12:04
From: Meggen Stockstill
Subject: Thank you (simplifying imaginary numbers)

Thank you for your prompt response.  I have at least two very talented 
students who were bothered by the apparent inconsistency, and while I 
can usually answer most of their questions, I wasn't sure if that 
product property was only for non-negative reals.  Our book does state 
that, but we weren't sure why it shouldn't hold for complex numbers as 
well.  It's been a long time since I studied the nitty gritty of 
complex numbers, planes, etc., so I appreciate your quick jogging of 
my memory. 

Thanks for your help,
Meggen Stockstill