Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone else has a similar issue.
Exercise 3, page 379 of "Basic Mathematics" by S.Lang.
Problem: Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$ $?
My approach to the question goes as follows:
Let $z = x + iy$ for real numbers $x, y$ different than $0$.
We have
\begin{align}
z\sqrt{z} & = (x + iy)\sqrt{z + iy} \\
& = x\sqrt{x+iy} + iy\sqrt{x+iy} &&\text{by distributivity}
\end{align}
We have
\begin{align}
|z\sqrt{z}| & = \sqrt{(x\sqrt{x + iy})^2 + (y\sqrt{x + iy})^2} &&\text{by definition} \\
& = \sqrt{x^2(x + iy)^2 + y^2(x + iy)^2} \\
& = \sqrt{(x^2 + y^2)(x + iy)} &&\text{by factoring} \\
& = \sqrt{(x^2 + y^2)} \sqrt{(x + iy)} \\
& = |z| \sqrt{z} &&\text{by definition} \\
\end{align}
The author's solution is $1$.
Thank you.
 A: The simpler way is to use the polar representation. For $z=\rho e^{i\theta}$ and using only one value of the square root we have:
$$
z\sqrt{z}=\rho e^{i\theta}\sqrt{\rho} e^{i\theta/2}=\rho\sqrt{\rho}e^{i3\theta/2}
$$
so $|z\sqrt{z}|=\rho\sqrt{\rho}=|z|\sqrt{|z|}$ and it is $=1$ only if $|z|=1$.
A: https://ernstchan.com/b/src/1457375466-129.pdf
The question is actually "what is the value of $z /\overline{z}$.
Which is easy.  By theorem 2:
$|z /\overline{z}| = |z|/|\overline{z}| = \sqrt{x^2 + y^2}/\sqrt{x^2 + (-y)^2} = 1$
;I don't see if Lang ever stated this in the text but it should be obvious $|z| = |\overline z|$
A: Using polar coordinates, you write $z=re^{i\theta}$, from which $\bar z=re^{-i\theta}$, thus
$$
z/\bar z=\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}
$$
whose absolute value is $1$ as requested.
A: Let $w=z\sqrt{z}$ (where $\sqrt{z}$ is any determination); then
$$
w^2=z^3
$$
and so
$$
|w^2|=|z^3|
$$
which implies
$$
|w|^2=|z|^3
$$
Therefore
$$
|w|=\sqrt{|z|^3}=|z|^{3/2}
$$

For the question about the absolute value of $z/\bar{z}$, consider that
$$
|z/\bar{z}|=|z|/|\bar{z}|
$$
and $|\bar{z}|^2=\bar{z}\,\bar{\bar{z}}=z\bar{z}=|z|^2$, so $|z|=|\bar{z}|$.
