Talk:Absolute Value of Power
Methinks 'twould be more valuable to prove the version for a positive real base and real exponent without reference to complex numbers and using "absolute value", and to prove the more general form (complex base; real exponent) as well. --Dfeuer (talk) 02:12, 11 April 2013 (UTC)
- Play with it as you'd like, I'm all proofed out right now. --GFauxPas (talk) 02:14, 11 April 2013 (UTC)
How about this (+ links)? --Linus44 (talk) 11:33, 13 April 2013 (UTC)
- Sounds like a good idea. Name theorem 2 something like Modulus of Power? The definition of argument does not seem to pin down $\operatorname{arg}$ very well. --Dfeuer (talk) 17:32, 13 April 2013 (UTC)
- That should not be a reason not to use the term "argument" - it just means that nobody's got round to attempting to define it rigorously yet. --prime mover (talk) 18:23, 13 April 2013 (UTC)
- Since Linus44 appears to know much more about complex analysis than I do, perhaps he can do so. But it's also possible that $\operatorname{arg}$ is inherently a bit tricky to pin down (perhaps having a codomain consisting of equivalence classes and used only in formal equations? I really don't know) so it may be easier to use $\operatorname{Arg}$ instead where appropriate. --Dfeuer (talk) 18:41, 13 April 2013 (UTC)
- He probably does. --prime mover (talk) 19:21, 13 April 2013 (UTC)
Theorem 1
Let $z \in \R_{\geq 0}$, $w \in \R$.
Then $|z^w| = |z|^w$.
Proof
Since $z \in \R_{\geq 0}$, $|z| = z$, so it is sufficient to prove that
- $z^w = |z^w|$
This is true if and only if $z^w \geq 0$.
But
- $z^w = \exp(w \log z)$
Since $z \geq 0$, $\arg(z) = 0$, so by Definition:Logarithm $\log z \in \R$.
Now by Exponential on Real Numbers is Group Isomorphism shows that $\exp(w \log z) = z^w \geq 0$.
Theorem 2
Let $z,w$ be complex numbers.
Then $|z^w| = |z|^w$ if and only if either $z \in \{0,1\}$ or $\Im(w) = 0$.
Proof
Step 1: $ \Leftarrow $
The statement is clear when $z \in \{0,1\}$.
Therefore suppose that $z \neq 0$ and $\Im(w) = 0$.
We have the definition the general complex power:
- $z^w = \exp(w\log z) = \exp(w \log |z| + i w\arg(z))$
so by Modulus and Argument of Complex Exponential,
- $|z^w| = \exp(\Re(w) \log |z| - \Im(w) \arg(z))$
Now, again by the definition of the general complex power and the complex logarithm,
- $|z|^w = \exp(w\log |z|) = \exp(\Re(w)\log |z| + i \Im(w) \log |z|)$
By Exponential on Real Numbers is Group Isomorphism $z \neq 0 \implies z^w \neq 0$, so we can write
- $|z^w|^{-1}|z|^w = \exp(i \Im(w) \log |z| - \Im(w) \arg(z))$
and by Exponential of Zero $|z^w|=|z|^w$ iff
- $(1):\qquad i \Im(w) \log |z| - \Im(w) \arg(z) = 0$
Since by hypothesis $\Im(w) = 0$, this relation is satisfied by Ring Product with Zero, so we indeed have $|z^w|=|z|^w$.
Step 2: $\Rightarrow$
Suppose that $|z^w|=|z|^w$.
If $z = 0$, the result is clear.
Suppose therefore that $z \neq 0$.
We have from $(1)$:
- $|z^w|=|z|^w$ iff $i \Im(w) \log |z| - \Im(w) \arg(z) = 0$
If $\Im(w) \neq 0$, by Complex Numbers form Integral Domain this implies that
- $i \log |z| = \arg(z)$
Since by definition $\arg(z) \in \R$, and $\log |z| \in \R$, we can only have this equality if
- $\log |z| = \arg(z) = 0$.
Since $\log |z| = 0$, we must have $|z| = 1$.
Since $\arg(z) = 0$, we must have $z \in \R_{\geq 0}$.
Therefore $z = 1$.
Thus either $z \in \{0,1\}$ or $\Im(w) = 0$ as required.
Why is this proof in the discussion page and not on the main theorem? --GFauxPas (talk) 08:24, 9 December 2016 (EST)