Proving the 'cut plane' $\mathbb{C}_{\pi}$ is an open, path connected set I want to show the 'cut plane' $\mathbb{C}_{\pi}$ is a domain, where $\mathbb{C}_{\pi}:=\{z\in\mathbb{C}: z\neq x \in \mathbb{R}, x\leq0\}$. My text defines a domain as an open, path connected subset of the complex plane.
So let's first show $\mathbb{C}_{\pi}$ is open. Let $z_0\in \mathbb{C}_{\pi}$. We just have to show there exists an  $\epsilon>0$-neighborhood about $z_0$, $N_{\epsilon}(z_0),$ that is completely contained in $\mathbb{C}_{\pi}$. Note then that if $z_0=x_0+iy_0$ and $y_0=0$, then $x_0>0$ by definition. In light of this, let's consider two cases:
$y_0\neq0: $
Choose $\epsilon:=\frac{|y_0|}{2}$ (which we can do since $y_0$ is non-zero). Then we just have to show every point in the neighborhood $N_{\epsilon}(z_0)$ has nonzero imaginary part. If $|z-z_0|<\epsilon=|\Im(z_0)|/2\iff2|z-z_0|<|\Im(z_0)|\iff... \text{(here's where I'd like to show)}...\iff|\Im(z)|>0$. This would imply imply $N_{\epsilon}(z_0)\subset\mathbb{C}_{\pi}. $
$y_0=0:$
By definition, $z_0=x_0$ where $x_0>0\in\mathbb{R}$. Then we just have to show every point in the neighborhood $N_{\epsilon}(z_0)$ is completely contained in $\mathbb{C}_{\pi}$ i.e. has positive real part. Take $\epsilon:=\frac{x_0}{2}$ (which we can do since $x_0$ is always nonzero). Then $|z-z_0|<\epsilon\iff (x-x_0)^2+y^2<\frac{1}{4}x_0^2\iff...\text{(here's where I'd like to show)}...$ $\iff\Re(z)>0$. This would imply $N_{\epsilon}(z_0)\subset\mathbb{C}_{\pi}.$
This would show $\mathbb{C}_{\pi}$ is an open subset of the complex plane since every neighborhood of any point of $\mathbb{C}_{\pi}$ is completely contained in it.
Now we show $\mathbb{C}_{\pi}$ is path connected. To do this, take any two points $z_0=r_0e^{i\theta_0}$ and $z_1=r_1e^{i\theta_1}$ in $\mathbb{C}_{\pi}$ with $-\pi<\theta_0, \theta_1< \pi$ and show there exists a path in $\mathbb{C}_{\pi}$ from $z_0$ to $z_1.$
Take as our path $$\gamma(t)=\begin{cases} 
      r_0e^{it} & t\in[\theta_0,\theta_1] \\
      ((t+1)-\theta_1)r_0e^{i\theta_1} & t\in[\theta_1, \frac{r_1}{r_0}+\theta_1-1 ]
   \end{cases}$$
which is always in $\mathbb{C}_{\pi}$ since $\theta_0\leq\arg(\gamma(t))\leq\theta_1$. Note one may also elementarily check this path is continuous everywhere since in particular $\lim_{t\to\theta_1^-}=r_0e^{i\theta_1}=\lim_{t\to\theta_1^+}$ so that $\gamma(t)$ is continuous at $\theta_1$. One may also check for all other $t\in[\theta_0, \frac{r_1}{r_0}+\theta_1-1 ]$ that $\gamma(t)$ is continuous since $$\lim_{t<\theta_1}\gamma(t)=\gamma(t_0)\text{  and } \lim_{t>\theta_1}\gamma(t)=\gamma(t_0)$$
Thus since $\gamma(t)$ is a (continuous) path joining points $z_0$ and $z_1$ in $\mathbb{C}_{\pi}$ lying entirely in $\mathbb{C}_{\pi}$, we conclude that $\mathbb{C}_{\pi}$ is path connected.
Finally, since we've shown $\mathbb C_{\pi}$ to be an open subset of $\mathbb{C}$ and path connected, it is a domain. $\blacksquare$
I am getting stuck on how to show my chosen $\epsilon$'s force $N_{\epsilon}(z_0)$ to be in $\mathbb{C}_{\pi}$ for each case. Any hints would be appreciated!
 A: By Pythagoras,
$$|\Im z - \Im z_0|^2 + |\Re z - \Re z_0|^2 = |z - z_0|^2  $$
Since the second term is nonnegative,
$$  |\Im z - \Im z_0|^2 \leq |z - z_0|^2  $$
and since squaring is monotonic,
$$  |\Im z - \Im z_0| \leq |z - z_0|  \text{.}  $$
So for $z \in B_{|y_0|/2}(z_0)$,
$$  |\Im z - \Im z_0| \leq |z - z_0| < |y_0|/2 \text{.}  $$
Using usual triangle inequality manipulations,
\begin{align*}
|\Im z - \Im z_0| &< |y_0|/2  \\
\Im z_0 - |y_0|/2 &< \Im z < \Im z_0 + |y_0|/2  \\
y_0 - |y_0|/2 &< \Im z < y_0 + |y_0|/2
\end{align*}
$$
\begin{cases}
0 < y_0/2 < \Im z < 3y_0/2 ,& y_0 > 0 \\
3y_0/2 < \Im z < y_0/2 < 0,& y_0 < 0
\end{cases}
$$
In either case $|\Im(z)| > 0$.
For $y_0 = 0$, do the analogous thing for real parts; you can discard half since $x_0 > 0$ is already known.
Also, for path connectedness, much easier than spiralling from one point to the other (and fussing to get the correct interval of angles): choose a "waystation" point on the positive real axis.  (You can choose one point for all time or choose a new one each time you try to find a path for a pair of points.)  The line segments between each point and the waystation point are in $C_\pi$ and are a path from one to the other.
A: It’s open as its complement is $N =  \{z\in \Bbb C: z \in \Bbb R, z \le 0\}$ so the complement is closed.
It’s connected: let for each $r>0$: $C_r = \{z \in \Bbb C\mid |z|=r, z \notin N\}$ which is connected as a circle minus its one point on the negative real axis. $P=\{z \in \Bbb C\mid z \in \Bbb R, z>0\}$ is also connected and intersects each $C_r$.
So $\Bbb C_\pi=\bigcup \{C_r: r > 0\}\cup P$ is connected by standard theorems on connected sets. A connected open subset of $\Bbb C$ is automatically path-connected (as $\Bbb C$ is locally path-connected), also a standard topology fact.
