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:

$\left \vert {x^y} \right \vert = \left \vert {x} \right \vert ^y$


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 $r = \left \vert {x}\right \vert$ and $\theta$ is an argument of $x$.

Then:

\(\displaystyle x\) \(=\) \(\displaystyle r e^{i\theta}\)
\(\displaystyle \implies \ \ \) \(\displaystyle x^y\) \(=\) \(\displaystyle \left(r{e^{i\theta} }\right)^y\)
\(\displaystyle \) \(=\) \(\displaystyle r^y e^{i \theta y}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left \vert {x^y} \right \vert\) \(=\) \(\displaystyle \left \vert {r^y e^{i \theta y} } \right \vert\)
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {r^y} \right \vert \left \vert {e^{i \theta y} } \right \vert\) Modulus of Product
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {r^y} \right \vert\) Modulus of Exponential of Imaginary Number is One
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {\left \vert {x} \right \vert^y} \right \vert\) by definition of $r$
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {x} \right \vert^y\) as $\left \vert {x} \right \vert^y \ge 0$

$\blacksquare$