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I don't really understand why the absolute value of a complex number is defined as its magnitude.

The absolute value for a real number, I think has two sensible interpretations: the magnitude of this number from the origin or the positive value of this number.

When extrapolating this to complex numbers, why is the first definition chosen, and not simply that the absolute value of a complex number is a complex number in the first quadrant (of the complex plane).

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    $\begingroup$ For example, $|-a|=|a|$ is a nice propery, which doesn't hold in your case. $\endgroup$
    – ThePortakal
    Mar 4, 2017 at 22:13
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    $\begingroup$ You could define some operation reflecting a number across the axes to get it into the "positive" quadrant - but I'm not sure that would be really useful. Have you seen any applications of the absolute value? Definitions are usually best informed by their applications. $\endgroup$
    – Milo Brandt
    Mar 5, 2017 at 1:01

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In general, we invent definitions and notation because we intend to make use of them somewhere. For example, the term "absolute value" and the corresponding notation $|-|$ exist because we regularly have occasion to refer to it in e.g. the definitions of variance of a random variable, limit of a sequence, and other constructions. For every one of these applications, the corresponding concept in the complex numbers is captured by magnitude. In contrast, I cannot think of a single case where I have needed to refer to "the number in the first quadrant differing from this one by a factor of a power of $i$."

Moreover, the useful algebraic properties of the absolute value function on the reals are not true of the function you've described. For example, if we denote your function by $[-]$, it is not the case in general that $[xy] = [x][y]$ -- how could it be, since the first quadrant isn't even closed under multiplication?

The point is, you can define whatever function you want, but if it's just a curiosity and not something that comes up naturally then it's probably not worth endowing with its own special notation and terminology.

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  • $\begingroup$ Many points well taken. $\endgroup$
    – Lubin
    Mar 5, 2017 at 19:32
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I think you chose the wrong analogy.
How about this instead: rotate your given complex number by such an angle that the result is on the positive real axis. You can think of the real absolute value as doing just that, after all.

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    $\begingroup$ Well, the real reason that your suggestion is not so good is that the usual absolute value is a multiplicative function: for complex numbers $z$ and $w$, we have the ultrapleasing relation $|zw|=|z|\cdot|w|$. $\endgroup$
    – Lubin
    Mar 5, 2017 at 2:58
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"The positive value of the number" is the effect of taking the absolute value of a real number, not the definition of absolute value. The actual definition of absolute value is a number's distance from zero. This is why the absolute value of a complex number is a positive real number, because distance is a real number.

Secondly, you can't argue that a complex number in the first quadrant of the complex plane is positive, since the imaginary unit ($i$) is considered neither positive nor negative.

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  • $\begingroup$ but there are positive and negative imaginary numbers $\endgroup$
    – JobHunter69
    Mar 4, 2017 at 22:09
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    $\begingroup$ But $i$ by itself is neither positive or negative, so $ai$, if $a$ is real, is not considered positive nor negative. The sign (The characters $+$ or $-$) in front of the number doesn't say anything about the actual positivity or negativity or the number. $\endgroup$
    – tc216
    Mar 4, 2017 at 22:11
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    $\begingroup$ @Goldname: If $i$ is defined such that $i^2=−1$, then we can show that $(−i)^2=−1$ . . . at which point we discover that $−i$ actually satisfies our original definition for i. Since we can also show that $−i≠i$, this means our definition was ambiguous. There is in fact no unambiguous way to define $i$ that could distinguish it from $−i$; every property of the one is true of the other. So we just say that $-1$ has two square roots, which are each other's additive inverses, and that we arbitrarily call one $i$ and the other $−i$ rather than the reverse. $\endgroup$
    – ruakh
    Mar 5, 2017 at 1:37
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    $\begingroup$ @ruakh Still, you can do the same with reals. 1^2 = 1 and (-1)^2 = 1, but -1 != 1, so I'm still not really understanding your argument. $\endgroup$
    – JobHunter69
    Mar 5, 2017 at 3:19
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    $\begingroup$ @Goldname: No, you cannot do the same with reals, because we do not define $1$ as being the number such that $1^2 = 1$. There are obvious ways to distinguish $1$ and $-1$; for example, only the former is a solution of $x^2 = x$. By contrast, there is literally no way to distinguish $i$ and $-i$; they are distinct, but any statement that's true of the one is, mutatis mutandis, true of the other. $\endgroup$
    – ruakh
    Mar 5, 2017 at 3:32
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Think of the imaginary number i as a rotation. Since i^2 = -1 and i^4 = 1, the imaginary number does not change magnitude but does change direction, which is why the absolute value is one and the same. Moreover, say you want to find the magnitude of z = a + bi. It follows that magnitude z = sqrt(|z|^2). Since i and -i are rotations in opposite directions, combining both rotations should bring one back to their original location. For this reason, we can say |z|^2= |z* z| where z* is the conjugate of z.

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