# When is square root the inverse of the square?

For years I have been wondering about something, and this is the day when the problem shall be rectified forever, by the help of you, ofcourse! :) It seems no one I ask really know the answer.

Today I came across this problem, and in it, there was an expression like this:

$$\cdots = (1-\cos^2 \alpha)^{1/2} = \cdots = (\sin^2 \alpha)^{1/2} = ((\sin \alpha)^2)^{1/2} \stackrel{\text{?}}{=} \begin{cases} |\sin \alpha| & \text{or is this incorrect, and rather}\\\sin \alpha \end{cases}$$

Basically what I want to know is when $a \in \mathbb{C}$, under which conditions are $(a^2)^{1/2} = a$ and when is $(a^2)^{1/2} = |a|$.

Thank you for your time.

Kind regards,

Marius

• $a\in \mathbb K$? – user99914 Feb 18 '15 at 9:59
• What do you mean by $\mathbb{K}$? – Martigan Feb 18 '15 at 9:59
• If $a$ is real, then $\sqrt{a^2}=|a|$; so $\sqrt{a^2}=a$ if and only if $a\ge0$. End of the story. – egreg Feb 18 '15 at 10:01
• $(x^2)^{1/2}$ cannot be the identity function because $x^2=(-x)^2$ and you would get $x=-x$. – Yves Daoust Feb 18 '15 at 10:01
• I wrote a discussion at Why does $\sin^{-1}(\sin(\pi))$ not equal $\pi$ that may be helpful. Although the original question is about $\sin^{-1}(\sin(x))$, I discussed $\sqrt{x^2}$ and $\left(\sqrt x\right)^2$ at length as a simpler example of the same thing. – MJD Feb 18 '15 at 15:02

I guess you're talking about real numbers. The most common definition of $\sqrt{x}$ is:
for real $x$, $x\ge0$, $\sqrt{x}$ is the (only) real number $y$ such that $y\ge0$ and $y^2=x$.
It follows from this definition that $\sqrt{a^2}=|a|$ (try it).
Thus $\sqrt{a^2}=a$ if and only if $a\ge0$.
The two functions $a \mapsto a^{\frac{1}{2}}$ and $a \mapsto \sqrt{a}$ are the same function : let's call it $sq$. It has a precise definition: it is the inverse function of $$[0, +\infty) \longrightarrow [0, +\infty) \qquad x \mapsto x^2$$
So $sq$ is defined on non-negative real numbers and has non-negative values. In particular for all real numbers holds $\sqrt{a^2} = |a|$.