Sketch the set $\Omega$ in the complex $z$-plane and the image $f \left(\Omega \right)$ in the $f(z)$-plane. Let $\Omega$ be the unit disk $\left|z \right| < 1$. Sketch the sets $\Omega$ and $f \left(\Omega \right)$, where $f_1(z)=z-i$ and $f_2(z)=2z+3i$.
For $f_1$: $\Omega$ is the unit disk centered at $0$ in the (complex) $z$-plane. The set $f \left( \Omega \right)$ is the line segment connected by the endpoints $(-1-i)$ and $(1-i)$ (but not including these points) in the $f(z)$-plane.
For $f_2$: $\Omega$ is the same as above. The set $f \left( \Omega \right)$ is the line segment connected by the points $2+3i$ and $-2+3i$ (but with the points omitted) 


Are these the correct sketches?
 A: Not quite.  Your graph of $\Omega$ is correct, though.
Since $f_1(z) = z - i$, then $f_1(\Omega) = \{ z-i : z \in \Omega \}$.  Basically, $f_1(\Omega)$ takes the entire set $\Omega$ and shifts it by $i$ units.  Shifts it in which direction?  To understand that, we can think about what happens to the point $i/2 \in \Omega$.  We have $f_1(i/2) = i/2 - i = -i/2$.  So the point $i/2$ was shifted down $i$ units (where "$i$ units" really just means "$1$ unit in the imaginary [vertical] direction").  So then the graph of $f_1(\Omega)$ is the graph of $\Omega$ shifted down $1$ unit, i.e., it's the open disk of radius $1$ centered at $(0,-i)$.
Obtaining the picture for $f_2(z)$ will be similar.  Since $f_2(z) = 2z + 3i$, then $f_2(\Omega) = \{ 2z + 3i : z \in \Omega\}$.  Use the same reasoning as in $f_1$ but be a bit more careful because the function is slightly more complicated.  You should get a disk that's twice as large and shifted up $3i$ units, i.e., the open disk of radius $2$ centered at $(0,3i)$.
Let me know if you need more details.
