# arg(\frac{1}{z})

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• Sep 24th 2010, 04:10 PM
dwsmith
arg(\frac{1}{z})
$arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}$

$arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli ne{z}}{z\overline{z}}\right)$
$=arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o verline{z})-arg(|z|^2)$
$=-\theta-(\theta+2\pi k)=-2\theta-2\pi k$

Is this correct?
• Sep 24th 2010, 04:14 PM
mr fantastic
Quote:

Originally Posted by dwsmith
$arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}$

$arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli ne{z}}{z\overline{z}}\right)$
$=arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o verline{z})-arg(|z|^2)$
$=-\theta-(\theta+2\pi k)=-2\theta-2\pi k$

Is this correct?

Note that $\displaystyle \frac{1}{z} = \frac{\text{cis} (0)}{r \text{cis} (\theta)} = \frac{1}{r} \text{cis} (0 - \theta)$.
• Sep 24th 2010, 04:17 PM
dwsmith
So the the argument of 1 over z is just going to be $-\theta+2\pi k$?
• Sep 24th 2010, 04:29 PM
Plato
Let’s consider the principal argument, $-\pi < \text{Arg}(z)\le \pi$.
The $\text{Arg}(\overline z)$ is simply its reflection in the real axis.

You may want to reconsider?
• Sep 24th 2010, 04:34 PM
dwsmith
If we pick up from where Mr F stopped, we have the angle as $0-\theta=-\theta$.

Now, my question is then why isn't the $arg\left(\frac{1}{z}\right)=-\theta+2\pi k \ \mbox{?}$
• Sep 24th 2010, 05:06 PM
chiph588@
Quote:

Originally Posted by dwsmith
$arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}$

$arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli ne{z}}{z\overline{z}}\right)$
$=arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o verline{z})-arg(|z|^2)$
$=-\theta-(\theta+2\pi k)=-2\theta-2\pi k$

Is this correct?

Given $c\in\mathbb{R}$ and $z\in\mathbb{C}$ then $\text{arg}(cz)=\text{arg}(z)$.

What does this say about $\text{arg}\left(\frac{\overline{z}}{|z|^2}\right) \; ?$
• Sep 24th 2010, 05:09 PM
dwsmith
$arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\ov erline{z})=-\theta+2\pi k$, yes?
• Sep 24th 2010, 05:17 PM
chiph588@
Quote:

Originally Posted by dwsmith
$arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\ov erline{z})=-\theta+2\pi k$, yes?

Yes, since z is reflected about the x-axis.
• Sep 24th 2010, 05:18 PM
zzzoak
I think that
$
arg(|z|^2)=arg(real \; number)=0
$

$
arg(1/z)=arg(\frac{1}{r} \; e^{-if})=-f+2 \pi n
$