Finding the discontinuities of $(z^2 - 1)^{\frac{1}{2}}$ $$
z^2 - 1 = \begin{cases}|z^2 - 1|^{1/2}e^{i\mathrm{Arg}(z^2 - 1)/2}, \quad z\neq \pm 1 \\ 0 \qquad \qquad \qquad\qquad\qquad z = \pm 1\end{cases}
$$
where $\mathrm{Arg}$ is the principal argument function.  
This is what I tried:
$z^2 - 1$ is discontinuous whenever $\mathrm{Arg}$ is discontinuous.
$\mathrm{Arg}$ is discontinuous whenever $\Re(z^2 - 1)\leq 0$ and $\Im(z^2-1)=0$.
$z^2 - 1 = x^2 - y^2 - 1 + 2ixy$
So $\mathrm{Arg}$ is discontinuous when
$$x^2 - y^2 - 1\leq0 \quad \text{and}\quad 2ixy = 0\\ \iff x^2 -y^2 - 1\leq0 \quad\text{and}\quad x=0 \ \text{or} \ y=0 \\ \iff (x^2 - y^2 - 1 \leq 0 \ \text{and} \ x=0) \ \text{or} \ (x^2 -y^2 - 1 \leq 0 \ \text{and} \ y=0) \\ \iff (-y^2 - 1 \leq 0) \ \text{or} \ (-1 \leq x \leq 1)  
 $$  
which is always true since the first statement is true.  
Did I go wrong somewhere? This can't be discontinuous everywhere right?
 A: 
Answer to the Question: Since the map $z\mapsto z^2-1$ is continuous on $\Bbb C,$  and the domain of continuity of $\sqrt z$  is $$\color{blue}{\Bbb D_{\sqrt z} =\Bbb C\setminus (-\infty,0]=\{z\in \Bbb C  \mid z\not\in (-\infty,0] \} } $$ we conclude  that the Domain of continuity of $z\mapsto \sqrt{z^2-1}$ is given by 
  $$\color{blue}{\Bbb D_{\sqrt{z^2-1}} =\{z\in \Bbb C  \mid z^2-1\not\in (-\infty,0] \}} $$
  Given that: $z^2-1 = (x+iy)^2-1=x^2-y^2-1+2ixy$
  But $$x^2-y^2-1 +2ixy \in (-\infty,0] \Longleftrightarrow \begin{cases}xy=0\\and\\x^2-y^2-1\le 0\end{cases}\\ \Longleftrightarrow z\in \{z=x+iy\in \Bbb C  \mid xy=0 \}  \cap \{z=x+iy\in \Bbb C  \mid x^2-y^2-1\le 0 \}     $$
  Therefore, by complementary we have
  $$\color{red}{\Bbb D_{\sqrt{z^2-1}} =\{z=x+iy\in \Bbb C  \mid x^2-y^2-1> 0 \} \cup \{z=x+iy\in \Bbb C  \mid xy \neq 0 \}} $$
  CHECK THE EXPLANATORY DETAILS BELOW

Recall: 
For every non-zero complex number $z$ there exist precisely two numbers $w_i$ such that $w^2 = z$: we therefore have to make some convention in other to get a right (or conventional)  definition of the complex square root function: tt is the principal square root of z (defined below). 
Let $z=re^{i\varphi}$, where $|z| = r ≥ 0$ is the modulus of $z$, and $φ$ is the Argument of $z$ (that is the angle that the line from the origin to the point makes with the positive real (x) axis.) 

Convention on the Square root: In complex analysis, this value is conventionally written If 
  $$\color{red}{ z = r e^{ i φ} ~~~~ \text{with}~~~  − π < φ ≤ π , \tag{I}}$$
  then we define the principal square root of z as follows:
  $$\color{blue}{\sqrt z= \sqrt r e^{i\varphi/2}}$$

The principal square root function is thus defined using the nonpositive real axis as a branch cut. 

Definition: Domain of continuity of $\sqrt z:$
  The principal square root function is only continuous on 
  $$\color{blue}{\Bbb D_{\sqrt z} =\Bbb C\setminus (-\infty,0]=\{z\in \Bbb C  \mid z\not\in (-\infty,0] \} } $$
  (See the justification below on for further details check here:)

Proof since any negative real number can written as $-x = -1*x$ with $x>0$ it suffices to prove the discontinuity at $z=-1$. 
Let $z_n = (-1+\frac{1}{n})e^{2in\pi} \to -1~~~as n\to \infty$
But $$\sqrt{z_n} = (1-\frac1n)e^{i\pi}e^{in\pi}= \frac{(-1)^{n+1} (n-1)}{n}$$
Which does not converges. Hence $$\lim_{n\to\infty} \sqrt{z_n} \neq \sqrt{-1}$$ whereas,$\lim_{n\to\infty} {z_n} \neq {-1}$. This prove the discontinuity.
A: $w\mapsto \sqrt{w}$ is discontinuous for $w\in \{(-t,0): t\geq 0\}$.
So $z\mapsto w = z^2-1 \mapsto \sqrt{w}$ is discontinuous for all
$z$ that verifies: $z^2-1=-t$, with $t\geq 0$ or, equivalently
 $$ z = \pm \sqrt{1-t}, \; \; t\geq 0$$
The interval $t\in [0,1]$ maps onto the segment $[-1;1]$ from $-1$ to $1$ and 
$t\in [1,+\infty)$ maps onto the imaginary axis $i{\Bbb R}$. The domain of continuity is the complement: $${\Bbb C} \setminus \left([-1;1] \cup i{\Bbb R}\right).$$
Regarding your proof, as already explained by fredgoodman, your last line is a wrong manipulation of logic. Otherwise, it is fine.
