Absolute Value of Power
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Theorem
Let $x$, $y$ be real numbers.
Let $x^y$, $x$ to the power of $y$, be real.
Then:
- $\size {x^y} = \size x^y$
Proof
If $x = 0$, the theorem clearly holds, by the definition of powers of zero.
Suppose $x \ne 0$.
We use the interpretation of real numbers as wholly real complex numbers.
Likewise we interpret the absolute value of $x$ as the modulus of $x$.
Then $x$ can be expressed in polar form:
- $x = r e^{i\theta}$
where $r = \size x$ and $\theta$ is an argument of $x$.
Then:
\(\ds x\) | \(=\) | \(\ds r e^{i\theta}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^y\) | \(=\) | \(\ds \left(r{e^{i\theta} }\right)^y\) | |||||||||||
\(\ds \) | \(=\) | \(\ds r^y e^{i \theta y}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {x^y}\) | \(=\) | \(\ds \size {r^y e^{i \theta y} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {r^y} \size {e^{i \theta y} }\) | Modulus of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {r^y}\) | Modulus of Exponential of Imaginary Number is One | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\size x^y}\) | by definition of $r$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \size x^y\) | as $\size x^y \ge 0$ |
$\blacksquare$
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