Continuous complex functions. We are given with a map $g:\bar D\to \Bbb C $, which is continuous on $\bar D$ and analytic on $D$. Where $D$ is a bounded domain and $\bar D=D\cup\partial D$. Then $\partial(g(D))\subseteq g(\partial D).$(I already know, how to prove it).
I need two examples: 
a) First, to show that the above inclusion can be strict, that is: $\partial(g(D))\not= g(\partial D).$
b) Second example, I need to show that conclusion in (1) is not true if $D$ is  not bounded.

There is an example, I was working on yesterday. But I couldn't understand it completely.
a) If we take $g(z)= z^2$ and $D$ =\begin{cases}z, & \text{where 1<|z|<2} \\\end{cases}
This $g$ is not 1-1. 
Now, we want to prove that $g(\partial D)\not\subset \partial(g(D)) $. Therefore, we need to show that $\exists $ some $z\in g(\partial D)$ but $z \not\in \partial(g(D))$. 
How will we show that ???
I want to talk about domain $D$ and its image by map $g$. Please check it: 
??
 A: I am thinking about the following example for b):
Let $g(z) = \frac{1}{z}$ and $D = \{ z \in \mathbb{C} \mid |z| > 1 \}$.  Then $\partial D$ is the unit circle, i.e. $\{ z \in \mathbb{C} \mid |z| = 1 \}$, and therefore $g(\partial D) = \partial D$.  We also have that $g(D)=  \{ z \in \mathbb{C} \mid |z| < 1 \text{ and } z \neq 0 \}$.  But this means that $\partial g(D)= \{ z \in \mathbb{C} \mid |z| = 1 \} \cup \{ 0 \}$.  So we have that $0 \in \partial g(D)$ but  $0 \notin  g(\partial D)$.
This shows that $\partial g( D) \not \subset g(\partial D)$ if $D$ is not bounded.
To give an example for a) I slightly modify the one you proposed.  I define $D$ to be 
\begin{equation}
D= \{ r * e^{i \phi} \mid 0<\phi<\frac{\pi}{2};\ 0\leq r \leq 1 \} \cup \{ r * e^{i \phi} \mid \pi<\phi<\frac{5}{4}\pi;\ 0\leq r \leq 1 \}
\end{equation}
and $g(z) = z^2$.  So $D$ is a quarter unit disk plus an eighth of a disk circle.  The trick is that the boundary of the eighth unit disk gets "mapped into" the image of the quarter disk.  Take for example the point $z_0=\frac{1}{2}e^{i \frac{5}{4}\pi}$.  This point is in $\partial D$ and gets mapped to $\frac{1}{4}e^{i \frac{1}{2}\pi}$ under $g$.  But $g(D) = \{r * e^{i \phi} \mid 0<\phi<\pi;\ 0\leq r \leq 1\}$ which means that $\frac{1}{4}e^{i \frac{1}{2}\pi} \notin \partial g(D)$.
