Is this formula equal to 1?

Question

Is this formula equal to $1$?

$$\frac{\left| \eta \right| }{\sqrt{\eta } \sqrt{\eta ^*}}$$

where $\eta$ is complex. And if so, how can I prove that?

Answer 1

By definition for $\eta=a+bi\in \mathbb C$

$$|\eta|=\sqrt {a^2+b^2}$$

$$\eta\cdot \eta^* =(a+bi)(a-bi)=a^2+abi-abi+b^2=a^2+b^2=|\eta|^2\in \mathbb R$$

then what is true is that

$$\sqrt{\eta\cdot \eta^*}=|\eta|\in \mathbb R$$

but, as already pointed out, $\sqrt{\eta}\sqrt{\eta^*}\neq \sqrt{\eta\cdot \eta^*}$ since in general $\sqrt{\eta}$,$\sqrt{\eta^*}\in \mathbb C$.

Answer 2

It is true that $|\eta|=\sqrt{\eta\cdot\eta^*}$ (in some books, this is the definition of absolute value, and in others it's easy to prove by setting $\eta=a+bi$ and using the Pythagorean theorem). As such we have $$ \frac{|\eta|}{\sqrt{\eta\cdot\eta^*}}=1 $$ (As long as $\eta\neq0$.) However, $\sqrt{\eta}\sqrt{\eta^*}$ is a dubious notion. I personally prefer to say it's undefined, and nonsensical. I suggest you do the same. Using square roots with complex numbers is something that warrants much care, and it can go wrong in subtle ways. It is better to just avoid entirely.

Answer 3

Some authors define $\sqrt\bullet$ as the "principal value" of the square root, i.e. the one such that $\Re\sqrt z\ge 0$ for all $z\in\Bbb C$ and such that $\Im\sqrt w>0$ for all $w\in(-\infty,0)$. With said assumption:

$$\sqrt z=\begin{cases}\sqrt{\frac{\lvert z\rvert+\Re z}2}+i\sqrt{\frac{\lvert z\rvert-\Re z}2}&\text{if }\Im z\ge0\\ \sqrt{\frac{\lvert z\rvert+\Re z}2}-i\sqrt{\frac{\lvert z\rvert-\Re z}2}&\text{if }\Im z<0\end{cases}$$

Where the square roots on the RHS are the usual ones $[0,\infty)\to[0,\infty)$. With this convention, $\sqrt{z^*}=(\sqrt z)^*$ holds for all $z\notin (-\infty,0)$, whereas $\sqrt{z^*}=\sqrt z=-(\sqrt z)^*$ for $z\in(-\infty,0)$. This means that, with this definition, $$\frac{\lvert z\rvert}{\sqrt z\sqrt{z^*}}=\begin{cases}1&\text{if }z\notin (-\infty,0]\\ -1&\text{if }z\in(-\infty,0)\end{cases}$$

I agree with everyone on preferring $\sqrt{zz^*}$. Notice that, no matter the definition of a function $f:X\to\Bbb C$ such that $(f(z))^2=z$, if $X$ contains a negative real number $\alpha$, then for said real number $\alpha^*=\alpha$ and $\lvert\alpha\rvert=-\alpha$, thus forcing $\frac{\lvert \alpha\rvert}{f(\alpha)f(\alpha^*)}=\frac{-\alpha}{(f(\alpha))^2}=-1$.