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### 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

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.

Meggen Stockstill
```
Associated Topics:
High School Imaginary/Complex Numbers
High School Square & Cube Roots

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