arg(\frac{1}{z})

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September 24th 2010, 03:10 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?
September 24th 2010, 03:14 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 $\displaystyle \frac{1}{z} = \frac{\text{cis} (0)}{r \text{cis} (\theta)} = \frac{1}{r} \text{cis} (0 - \theta)$ .
September 24th 2010, 03:17 PM
dwsmith

So the the argument of 1 over z is just going to be $-\theta+2\pi k$ ?
September 24th 2010, 03:29 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?
September 24th 2010, 03:34 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 $arg\left(\frac{1}{z}\right)=-\theta+2\pi k \ \mbox{?}$
September 24th 2010, 04:06 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 $\text{arg}\left(\frac{\overline{z}}{|z|^2}\right) \; ?$
September 24th 2010, 04:09 PM
dwsmith

$arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\ov erline{z})=-\theta+2\pi k$ , yes?
September 24th 2010, 04:17 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.
September 24th 2010, 04:18 PM
zzzoak

I think that
$<br /> arg(|z|^2)=arg(real \; number)=0<br />$

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