For $z \in \mathbb{C}\setminus \{0\}$, is $|1/z| = 1/|z|$? I have been trying to prove that for $z \in \mathbb{C}$ and $z \neq 0$ that $|1/z| = 1/|z|$ but with no success.  Not all the properties of real numbers can be applied to complex numbers, of course.
 A: Yes.
Note that $1/|z|$ is, by definition, the real multiplicative inverse of $|z|$, so by uniqueness of such an inverse, since
$$|1/z|\times |z|=|z/z|=1,$$
we have $|1/z|=1/|z|$.
A: Try to prove that $\overline{\frac 1 z} = \frac 1{\overline z}$.
Then $|\frac 1z| = \sqrt{\frac 1z\cdot \overline{\frac 1z}}=$
$\sqrt{\frac 1 z\cdot \frac 1{\overline z} } =\sqrt{\frac 1{z\overline z}}=$
$\sqrt{\frac 1{|z|^2}} = \sqrt{(\frac 1{|z|})^2}=$
$|\frac 1{|z|}| = \frac 1{|z|}$.
(bear in mind you do know that $|z|$ is a positive real number)
That's probably way more detain than you need.
.....
Alternatively
$|\frac 1{z}| = |\frac 1z\cdot \frac {\overline z}{\overline z}|=$
$|\frac {\overline z}{z\overline z}|= |\frac {\overline z}{|z|^2}|=$
$\frac 1{|z|^2}|\overline z|=\frac 1{|z|^2}|z|=$
$\frac 1{|z|}$.
Bear in mind $|z| = |\overline z|$ and $|z|^2 = z\overline z \in \mathbb R^+$ and $|z|\in \mathbb R^+$.
A: Suffices to show $|z|\cdot |1/z|=1.$ To do this, use that law that for $w,z\in \mathbb{C},$
$$
|z|\cdot |w|=|z\cdot w|.
$$
A: Maybe the easiest way to see that is to use exponential form of the complex number $z = \rho e^{i\phi}$. Thus:
$$
\frac{1}{z} = \frac{1}{\rho} e^{-i \phi} \implies \left|\frac{1}{z}\right| = \frac{1}{\rho}
$$
On the other hand:
$$
|z| = \rho \implies \frac{1}{|z|} = \frac{1}{\rho}
$$
A: The main problem that you have here is that you have been trying to re-invent the wheel, rather than actually consulting a Complex Analysis textbook.  The following results are based on Chapter 1 of "An Introduction To Complex Function Theory" [Bruce Palka].

*

*For $z,w \in \mathbb{C}, |z| \times |w| = |z \times w|.$


*For $z = (x + iy) \in \mathbb{C}$, let $\overline{z}$ denote $(x - iy).$


*Then $\frac{1}{z} = \frac{\overline{z}}{z \times \overline{z}}.$


*$z \times \overline{z} = |z|^2.$


*$|\overline{z}| = |z|$.
The result that you are trying to prove is a direct consequence of the above intermediate results.
