# Absolute Value of Power

## 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 $\left \vert {x} \right \vert^y \ge 0$ |

$\blacksquare$