Proof for argument identity I have trouble algebraically show proof for this well known statement:
$$-\operatorname{arg}(z)=\operatorname{arg}(z^{-1})=\operatorname{arg}(\bar{z})$$
if $z=x+yi$ and $-\operatorname{arg}(z)=-\arctan(y/x)$ right? This one I understood if $\arctan(-\phi)=-\arctan(\phi)$
$$\begin{align} -\arg (z)&=\arg(\bar{z}) \\ -\arctan(y/x)&=\arctan(-y/x) \\ -\arctan(y/x)&=-\arctan(y/x)\end{align}$$
But I have trouble with showing that $-\operatorname{arg}(z)=\operatorname{arg}(z^{-1})$ or is it something like this.
$$1/z=\frac {1}{x+iy}=\frac{x-iy}{x^2+y^2}$$
EDIT: I think I understood..
$$-\arg(z)=\arg(1/z)=\arctan \Big( \frac{\dfrac{-y}{x^2+y^2}}{\dfrac{x}{x^2 +y^2}} \Big)=\arctan(-y/x)=-\arg(z) $$
Did I get it right?
 A: So, because we're only considering the argument we can stay on the unit circle. Then, if $z = e^{i \theta}$ then $z^{-1} = e^{-i\theta} = \overline{z}$. 
Moreover
\begin{equation}
-\operatorname{arg}(z) = -\theta = \arg(z^{-1}) = \arg(\overline{z})
\end{equation}
A: Since you are talking about a non-zero complex number $z$, you should know that $z$ can always be written as $z=re^{i\theta}$ with $r>0$ and $\theta\in \mathbb R$.
This is a polar expression of $z$ and any other polar expression of $z$ is of the form $\rho e^{i\psi}$ where $\rho=r$ and $\psi=\theta+2k\pi$, $k\in \mathbb Z$.
Now, by definition, $arg(z)$ is given by $\theta$ up to $2\pi$ that is : $$arg(z)=\theta \pmod {2\pi}.$$
If the above explanation (or definition) is clear to you, you can now compute anything you want.
Observe that $$\frac{1}{z}=\frac{1}{r}e^{-i\theta}$$ and this is a polar expression so $$\begin{array}{rl} arg(z^{-1}) & =-\theta \pmod {2\pi} \\ & = -arg(z)\end{array}$$
Similarly, $$\bar{z}=\overline{re^{i\theta}} =\bar{r}\cdot \overline{e^{i\theta}}=re^{-i\theta}$$ which is a polar form, so $$\begin{array}{rl} arg(\bar z) & =-\theta \pmod {2\pi} \\ & = -arg(z)\end{array}$$
A: Using the inverse tangent function $\arctan$ to find the argument only works if you restrict yourself to two adjacent quadrants. For example, $1+\mathrm{i}$ and $\mathrm{-1}-\mathrm{i}$ both have 
$$\frac{y}{x}=1$$
and so would have the same argument if $\arg(z) = \arctan\!\left(\frac{y}{x}\right)$. 
Similarily, $-1+\mathrm{i}$ and $1-\mathrm{i}$ both have $\frac{y}{x}=-1$ and would have the same argument.
I would recommend that you use the polar form $z=r(\cos \theta + \mathrm{i}\sin\theta)$,
where $r$ is the modulus of $z$ and $\theta$ is the argument. The sine and cosine functions repeat every $2\pi$. We have:
$$\frac{1}{z} = \frac{1}{r(\cos\theta+\mathrm{i}\sin\theta)} = \frac{1}{r}\cdot\frac{1}{\cos\theta+\mathrm{i}\sin\theta}$$
We multiply numerator and denominator by the conjugate:
$$\begin{eqnarray*}
\frac{1}{z} &=& \frac{1}{r}\cdot\frac{1}{\cos\theta+\mathrm{i}\sin\theta}\cdot\frac{\cos\theta-\mathrm{i}\sin\theta}{\cos\theta-\mathrm{i}\sin\theta} \\ \\
&=&\frac{1}{r}\cdot\frac{\cos\theta+\mathrm{i}\sin\theta}{\cos^2\theta+\sin^2\theta} \\ \\
&=& \frac{1}{r}\cdot (\cos\theta-\mathrm{i}\sin\theta)
\end{eqnarray*}$$
Two well-known identities are: $\cos(-\theta) \equiv \cos \theta$ and $\sin(-\theta) \equiv - \sin\theta$. It follows that
$$\frac{1}{z} = \frac{1}{r}\cdot (\cos\theta-\mathrm{i}\sin\theta) = \frac{1}{r}\cdot(\cos(-\theta)+\mathrm{i}\sin(-\theta))$$
It follows that 
$$\begin{eqnarray*}
\left| \frac{1}{z} \right| &=& \frac{1}{|z|} \\ \\
\arg\left(\frac{1}{z}\right) &=& -\arg z
\end{eqnarray*}$$
