Absolute Value and Exponents

Question

In my homework I've been accustomed to assuming that $|x|^a = |x^a|$ Recently however, I've begun to doubt that. Take the following example:

$$ \begin{equation*} \begin{split} |\sqrt{-|x|} | &= \sqrt{|-|x||} \\ &= \sqrt{|x|} \end{split} \end{equation*} $$

I don't, it just seems weird to me. So I took it upon myself to investigate my assumption, and prove/ disprove that $|x|^a = |x^a|$.

$$\begin{equation*} \begin{split} |x|^a &= \begin{cases} x^a, \ \ & x \geq 0 \\ (-x)^a, \ \ & x < 0 \end{cases}\ \\[10pt] &= \begin{cases} x^a, \ \ & x \geq 0 \\ (-1)^a \cdot (x)^a, \ \ & x < 0 \end{cases}\ \\ \end{split} \end{equation*}$$

Whereas $$|x^a|= \begin{cases} x^a, \ \ & x \geq 0 \\ -x^a, \ \ & x < 0 \end{cases}\ $$

In other words, it would seem to me that the two only equal each other when $a$ is odd.

I don't know, could you guys clear this up for me?

Answer 1

It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.

Answer 2

Let's go a step further and look at this from the complex point of view. Any number can be written as:

$$z=r\cdot e^{i\theta}$$

with $r\in\mathbb{R}$ and $r\ge0$ and $\theta\in[0,2\pi[$.

raising this number to any power $a\in\mathbb{R}$ gives you

$$z^a=r^a\cdot e^{i\,a\,\theta}$$

The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $a\in\mathbb{R}$

For a complex power $a+bi\in\mathbb{C}$, this is not the case:

$$z^{a+bi}=r^{a+bi}\cdot e^{i(a+bi)\theta} = r^{a}e^{-b\theta}\cdot e^{i(\ln r + a\theta)}$$

which gives $|z^{a+bi}|=r^{a}e^{-b\theta}$ while $|z|^{a+bi}=r^{a+bi}$