# Absolute Value of Power

## 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 $\size x^y \ge 0$

$\blacksquare$