# Absolute value of imaginary numbers?

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• 08-09-2008
cpjust
Absolute value of imaginary numbers?
Hi,
Here's something that stumped my pre-calculus teacher.
Supposedly, the absolute value of an imaginary number such as 4 + 3i is found by throwing away the i and using the pythagorean theorem on it. i.e. sqrt( 4^2 + 3^2 ) = 5

So according to that formula, if you have 3 numbers:
x = 4 + 3i
y = 3 + 4i
z = 0 + 5i

Then,
| x | = | y | = | z | = 5

But if that is true, then the following should also be true, but it isn't:

Code:

```    x^2        =      y^2        =  z^2   (4 + 3i)^2    =    (3 + 4i)^2    =  (5i)^2 (16 + 24i - 9)  =  (9 + 24i - 16)  =  (-25)   7 + 24i      !=    -7 + 24i    !=  -25```
Part of the reason why I think that the formula is invalid is because they ignore the i which is part of the number.
The only "proof" that I've seen is graphical, but since they are dealing with imaginary numbers, how can they know that the imaginary number line is perpendicular to the real number line, or that the points on the imaginary number line have the same spacing as real numbers or even if it has a linear progression? If we knew the properties of imaginary numbers, they wouldn't be very imaginary. ;)
• 08-09-2008
tabstop
I can think of no earthly reason why anyone would think that |x| = |y| implies that x^2 = y^2. All it implies is that x*xbar = y*ybar (where xbar is the complex conjugate of x).

And yes, if you interpret imaginary numbers in the graphical way of radius+angle, what x^2 does is square the distance and double the angle. So your three x^2, y^2 and z^2 are all still of the same lengths, but the angles are (still) not the same.
• 08-10-2008
PING
An imaginary number can be considered as a vector, and a vector has magnitude as well as direction. You are going wrong in the first step itself. When you say |x| = |y|, it doesn't imply that x^2 = y^2 simply because the absolute value has no direction, it is a scalar and x^2 is a vector.
• 08-10-2008
Stonehambey
Quote:

Originally Posted by tabstop
I can think of no earthly reason why anyone would think that |x| = |y| implies that x^2 = y^2.

lol, well I can think of one: a somewhat hazy understanding of complex numbers. Illustrated below (not that it's a bad thing)

Quote:

Originally Posted by cpjust
how can they know that the imaginary number line is perpendicular to the real number line, or that the points on the imaginary number line have the same spacing as real numbers or even if it has a linear progression? If we knew the properties of imaginary numbers, they wouldn't be very imaginary. ;)

"Imaginary" is just a name, so try not to take it too literally. We know the properties of complex numbers because we define them.

A complex number, z, is defined as

z=a+bi

where a and b are in the set of real numbers, and i is the square root of minus one. So from this we construct what is known as the complex plane, which is the graphical representation you are speaking of. :)
• 08-10-2008
brewbuck
Quote:

Originally Posted by cpjust
The only "proof" that I've seen is graphical, but since they are dealing with imaginary numbers, how can they know that the imaginary number line is perpendicular to the real number line, or that the points on the imaginary number line have the same spacing as real numbers or even if it has a linear progression? If we knew the properties of imaginary numbers, they wouldn't be very imaginary. ;)

The numbers are "imaginary" because that's the term chosen by the guy who first started playing with the concept sqrt(-1). The "number line" itself does not actually exist, nor does the "complex plane." These are just concepts which allow us to visualize.

The most important thing to understand about complex numbers is that they are made up of two components which are independent of each other. The sqrt(-1) thing just adds some spice to the mathematics when these quantities multiply with each other -- in most other respects, complex numbers are very similar to 2-vectors. The best way to represent, and think, about this is in terms of two dimensions. As there are multiple representations of multidimensional quantites, so there are multiple representations of complex numbers. You can look at them in terms of Cartesian coordinates on a plane, or a magnitude and an angle from "reality."

Forget about the mysticism implied by the name "imaginary." It was a dumb name but unfortunately it has stuck. These numbers are quite REAL, at least so much as any other number you've ever encountered is "real," and play important roles in explaining physical phenomena.
• 08-10-2008
cpjust
Quote:

Originally Posted by cpjust
If we knew the properties of imaginary numbers, they wouldn't be very imaginary. ;)

This was actually meant as more of a joke.

But what I don't understand is why people assume that sqrt( -1 ) is perpendicular to the real number line?
• 08-10-2008
tabstop
Quote:

Originally Posted by cpjust
This was actually meant as more of a joke.

But what I don't understand is why people assume that sqrt( -1 ) is perpendicular to the real number line?

It depends on what you mean by "assume". Why do people assume that the y-axis is perpendicular to the x-axis? (I mean, you can get a perfectly good 2-d system with the y-axis at any angle (edit: other than 0 or 180 of course).) It's just more convenient to be orthogonal than not.
• 08-10-2008
laserlight
If I should hazard a guess, it is because a change in the real number coefficient (or the imaginary number coefficient) has no effect on the imaginary number coefficient (or real number coefficient, respectively), thus suggesting that they can be represented orthogonally.

EDIT:
hmm... the convenience argument sounds better, though then an argument might be put forth if we merely find what we are familiar with convenient.
• 08-10-2008
cpjust
Quote:

Originally Posted by tabstop
I can think of no earthly reason why anyone would think that |x| = |y| implies that x^2 = y^2. All it implies is that x*xbar = y*ybar (where xbar is the complex conjugate of x).

Well I couldn't use: (|x|)^2 because then how do I get rid of the | | without using the very formula I'm trying to disprove?

|x| is just trying to make x positive, so squaring a positive or negative number creates a positive number. Of course squaring i gives you a negative number, so maybe I should use ^4 instead? :D
The main reason I started playing around with this idea is because I've never seen an algebraic proof for the absolute value of imaginary numbers; only the graphical one and I just can't see why the i isn't included in the formula.
• 08-10-2008
tabstop
Quote:

Originally Posted by cpjust
Well I couldn't use: (|x|)^2 because then how do I get rid of the | | without using the very formula I'm trying to disprove?

|x| is just trying to make x positive, so squaring a positive or negative number creates a positive number. Of course squaring i gives you a negative number, so maybe I should use ^4 instead? :D
The main reason I started playing around with this idea is because I've never seen an algebraic proof for the absolute value of imaginary numbers; only the graphical one and I just can't see why the i isn't included in the formula.

I think you've missed the boat completely. |x| is defined as x*xbar, not as "making x positive" or any such -- it finds the length of x.
• 08-10-2008
Sang-drax
The formula
|a + ib| = sqrt(a^2 + b^2)
is true by definition. There is no proof. Given this definition, the fact that the "imaginary axis is perpendicular to the real axis" follows from this.
• 08-10-2008
Sang-drax
By the way, your question is a good one. It is important to keep track of what is defined and what is derived.
• 08-10-2008
cpjust
Quote:

Originally Posted by tabstop
I think you've missed the boat completely. |x| is defined as x*xbar, not as "making x positive" or any such -- it finds the length of x.

I've heard of foobar, but I've never heard of xbar? What is it?

In elementary school, you found the absolute value by removing the negative sign if it existed; in high school it became sqrt( x^2 )... Why don't they just tell us the real formula right from the beginning? :rolleyes:
• 08-10-2008
Thantos
Because sometimes you need simplifications before you can get to the truth. Imagine if you tried to get 5th graders to find the area of a triangle by integration instead of just giving them the derived equation?
• 08-10-2008
tabstop
Quote:

Originally Posted by cpjust
I've heard of foobar, but I've never heard of xbar? What is it?

In elementary school, you found the absolute value by removing the negative sign if it existed; in high school it became sqrt( x^2 )... Why don't they just tell us the real formula right from the beginning? :rolleyes:

Quote:

Originally Posted by tabstop
(where xbar is the complex conjugate of x).

Because most elementary school students aren't up on their inner product spaces. (If it comes to that, we haven't gotten to the real definition of | | in this thread yet either -- all we've gotten to are special cases of the general formula. And for that matter, there are an infinite number of norms (this is the 2-norm, but there's also the 1-norm and the 3-norm and the ... and the infinity-norm). See article.
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