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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.

Shaun
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El antonio banderas
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    This should follow from $|zw|=|z||w|$ and $|1|=1$. – Dave Jan 14 '21 at 22:35
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    note $|z||1/z|=|z\times 1/z|=|1|=1$ – J. W. Tanner Jan 14 '21 at 22:35
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    You should be able to show that $|z/w|=|z|/|w|$ in general. But in your case, just let $z=x+iy$, where $x,y\in\mathbb{R}$ and are not both zero, and directly evaluate both sides. – mjqxxxx Jan 14 '21 at 22:36
  • Not all the properties of real numbers might be applied to complex numbers, of course... still, $1/|z|$ is a real number. :^) – Ottavio Jan 14 '21 at 22:47

5 Answers5

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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 $|1/z|=1/|z|$.

Shaun
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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 $|z|\in \mathbb R^+$.

fleablood
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Suffices to show $|z|\cdot |1/z|=1.$ To do this, use that law that for $w,z\in \mathbb{C},$ $$ |z|\cdot |w|=|z\cdot w|. $$

subrosar
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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: $$ |z| = \rho \implies \frac{1}{|z|} = \frac{1}{\rho} $$

RedGiant
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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.

user2661923
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