1. ## 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?

2. 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 $\displaystyle \frac{1}{z} = \frac{\text{cis} (0)}{r \text{cis} (\theta)} = \frac{1}{r} \text{cis} (0 - \theta)$.

3. So the the argument of 1 over z is just going to be $-\theta+2\pi k$?

4. 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?

5. 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 $arg\left(\frac{1}{z}\right)=-\theta+2\pi k \ \mbox{?}$

6. 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 $\text{arg}\left(\frac{\overline{z}}{|z|^2}\right) \; ?$

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

8. 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.

9. I think that
$
arg(|z|^2)=arg(real \; number)=0
$

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