0

There are numbers which are square roots of negative numbers and they're are those which are logarithms of negative numbers(both are complex numbers).

Are there any numbers whose moduli are negative? I mean, $|x|<0$

If yes, what are they?

If no, why?

  • Arguably there are complex numbers $z$ where $z^2$ has a negative real part and a zero imaginary part, and there are complex numbers $z$ where $e^z$ has a negative real part and a zero imaginary part. So you may need to find a function on the complex numbers corresponding to the inverse of taking the absolute value – Henry Aug 09 '17 at 07:45
  • If you allow numbers with negative moduli, then I would say your modulus function doesn't deserve that name any more. – Arthur Aug 09 '17 at 08:16

3 Answers3

1

The modulus, by definition, is a positive number as it represents a concept similar to the "magnitude" in Physics. So the fact that the value of a modulus is always positive does not depend on the complex number you are considering at all but is in fact an attribute of the modulus itself.

In $\mathbb C$, given a complex number $x = a + ib$ (with $a$, $b ∈ \mathbb R$) the modulus is defined as

$$\sqrt{x² + y²} $$

Since $y$ itself real, the number under the square root is always positive, hence the modulus is always real and positive.

From a more general and abstract point of view, the modulus is a special case of the norm in $\mathbb C$. By definition, for any vector space, the norm satisfies the below conditions:

  • Must be real and positive
  • Separation
  • Absolute homogeneity
  • Subadditivity
Y2H
  • 161
1

The modulus is meant to represent the magnitude of a number, regardless its direction (also for the complex). This is precisely why it is defined to be a non-negative number.

You are free to define another "modulus", for example by keeping the ordinary modulus and doubling the argument:

$$[z]:=|z|e^{i2\angle z}=ze^{i\angle z}=\frac{z^2}{|z|}.$$

With this definition,

$$[1]=[-1]=1,\\ [i]=[-i]=-1,\\ [1\pm i]=[-1\pm i]=\pm i\sqrt2,\\\cdots$$

This is perfectly artificial.

  • Thanks, @YvesDaoust ! That's informative​!(+1) – Myungjin Hyun Aug 09 '17 at 08:24
  • I have to ask though. What is the incentive in calling this a "modulus" rather than say an "umspazzsneequanfloob". Or simply defining a modulus to mean $x^2 + e^x - 517$. I mean what conditions that define the concept of "modulus" does this satisfy? – fleablood Aug 09 '17 at 08:30
  • From Wikipedia: Modulus is the diminutive from the Latin word modus meaning measure or manner. It is a length. I would accept "umspazzsneequanfloob" for $x^2+e^x-517$. –  Aug 09 '17 at 09:11
1

In the complex numbers, no. $|z|$ is defined to be $\sqrt{z*\overline z}$ which is non negative as $z\overline z = \Re(z)^2 + \Im(z)^2$ is always non-negative real.

I suspect you are asking as we extended the reals to the complex (depending upon your philsophy) by allowing square roots of negatives which as a consequence allowed for logarithms of negative numbers, you are asking if we can create another number system that will allow $|z| < 0$.

It's hard for me to imagine how or why we would do so. There are no numbers that "want to exist but can't" because if they existed there modulus would have to be negative. And as modulus was defined to be non-neg real, it's hard to imagine in what sense we'd define another meaning for it that satisfy any condition associated with it. Primarily the condition that the modulus represents the quantitative measure of the absolute non-negative size of something.

=====

Okay, more.

In the reals $|x|$ is defined as $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

In extending to the complex numbers we could have kept that definition as close as possible by saying $|a+ib| = a+ib$ if $a \ge 0$ and $|a+ib| = -a-ib$ if $a < 0$. Admittedly we wouldn't ever have $|z| < 0$ but we would have $|z| \not \ge 0$. We could define, god knows why we'd want to but we could, $|a+ib| = a+b$ and the we could have $|z| < 0$.

So we don't we? Why instead to we replace the simple $|x| = \pm x$ with the scary looking $|a+bi| = \sqrt{(a+bi)(a-bi)} = \sqrt{a^2 + b^2}$?

Well, BECAUSE $|a+bi|= \sqrt{a^2 + b^2}$ IS greater than or equal to $0$. That $|z| \ge 0$ is a requirement of the definition of anything that we wish to call a "modulus".

Y2H in his/her answer lists some of the requirements for what something we call a "modulus" must obey. Why must we obey it to call it a modulus? Because if we didn't there wouldn't be any meaning to the word "modulus".

(Why is an elephant large, gray and wrinkly? Because if it were small, white and smooth it would be an aspirin.)

The very first requirement is ... that the "modulus" is real, and non-negative.

The modulus is essentially the "size" of a number. And size is positive value (or zero if and only if the number is zero). That's just.... axiomatic.

There are other conditions. By definition:

In an algebra, If $a,b \in F$, a field, $|a|$ is called the modulus, we must have:

i)$|a| \in \mathbb R; |a| \ge 0$.

ii) $|a| = 0$ if and only if $a$ is the multiplicative identity of $F$.

iii) $|ab| = |a||b|$

iv) $|a+b| \le |a| + |b|$

In vector spaces, $|x|$ is a norm and we must have (by definition) where $V,W$ are vectors in a vector space and $a$ is an element of a field:

i) $|V| \ge 0; |V| \in \mathbb R$

ii) $|V| = 0 \iff V = 0$

iii) $|aV| = |a||V|$ where $|a|$ is a modulus of $a$.

iv) $|V + W| \le |V| + |W|$.

fleablood
  • 124,253
  • Thanks, @fleablood ! You got the point of my question exactly! Well, I guess it's not possible to have such a number system! Thanks again! (+1) – Myungjin Hyun Aug 09 '17 at 08:27
  • Weren't logarithms too initially defined to be real (I'm not 100% sure on that), @fleablood , and later allowed to be complex? So why is the modulus different in that we cannot have a negative modulus? Just wondering! – Myungjin Hyun Aug 09 '17 at 08:32
  • I wouldn't say it is impossible. I'm just saying I don't see what conditions a modulus what have that would allow us to be "justified" in calling the result $|x|$. I mean we could define $|x| = x - 27$ for all we care. It's just that the is no justification for it. I don't know there is no way or reason for such a system, but I just don't see it. – fleablood Aug 09 '17 at 08:33
  • OK, thanks, @fleablood ! So my assumption is not correct with the current definition of modulus? I mean, one doesn't have to mess with the definition of a square root to get complex numbers! – Myungjin Hyun Aug 09 '17 at 08:39
  • $\log_k x$ "means" the power we must raise k to to get $x$. In the reals we can't raise a positive to a power and get a negative. And a negative base has to many undefined values to be of use. Allowing e^x to be so that $\frac{de^x}{dx} = e^x$ the discovery of complex allowed the extended def. What does $|x|$ "mean"? And what needs extending. $|x|$ means the magnitude of something. There may be some physics where magnitude is negative. But... I don't know. I don't see the sense in it. – fleablood Aug 09 '17 at 08:42
  • No. But we messed with the definition of multiplication. And of number. – fleablood Aug 09 '17 at 08:43
  • You mean, @fleablood , that we messed with the rule √(ab) =√a√b, right? I agree with your point! – Myungjin Hyun Aug 09 '17 at 08:58
  • Some rules are are observable consequences and other rules prescribed definitions. Multiplication obeys the field axioms and on real numbers, the ordered field axioms, and |x| is a positive magnitude are prescribe definitions. They can not be messed with. That $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and "negatives don't have square roots" are observed consequences. We "messed" with $ab=$ whatever ordered* fields allow, to $(a+ib)(c+id)=(ac-bd)+i(ad+bc)$ and we were only "allowed" to because it obeys the field axioms (but not ordered). .... to be continued.... – fleablood Aug 09 '17 at 15:56
  • ... the so called rules $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and "negatives don't have square roots" fell by the wayside and were not true. $|x| = \pm x$ was a rule no longer was acceptable simply because it wasn't positive *anymore. So $|a+ib|=\sqrt{(a-ib)(a+ib)}$ was "messed with". But we can't mess with $|x| \ge 0$ because that is the prescriptive definition. – fleablood Aug 09 '17 at 16:02