Find the image of the Möbius transformation My question is:
Find the image of the area $C=\{z\in \mathbb{C}:|z+3|\geq 3\}$ of the Möbius transformation $w=f(z)=\frac{z}{z+6}$.
I have drawn the image in the z-plane and then taken three points $z_{1},z_{2},z_{3}$ and then used the function for $w$ to get values for $w_{1},w_{2},w_{3}$. Then I have drawn the image for the w-plane. See figure. Is this correct?

 A: As HajJafar says, Mobius transformations map generalised circles to generalised circles. (A "generalised circle" is anything that is either a circle or a straight line.)
A generalised circle can be uniquely specified by specifying three points on it.
For example, the circle $C = \{z \in \mathbb C : |z + 3 | = 3 \} $ is the unique circle containing the three points $-3 +3i$, $0$ and $-3 - 3i$.
You have identified that your Mobius transformation maps $-3 +3i$, $0$ and $-3 - 3i$ to $i$, $0$ and $-i$ respectively. So the Mobius transformation must map the circle $C$ to the unique generalised circle passing through $i$, $0$ and $-i$.
The unique generalised circle passing through $i$, $0$ and $-i$ is easy to identify. It is the $y$-axis, $\{ z \in \mathbb C : {\rm Re}(z) = 0\}$.
[This is not the same thing as the line segment between $-i$ and $i$, which is what you drew on your picture.]
But you're asked to find the image of the region $\{z \in \mathbb C : |z + 3 | \geq 3 \}$. Observe that the point $z = 1$ is in this region, and it maps to the point $\frac 1 7$, which is on the right-hand side of the $y$-axis. The conclusion is that the image of $\{z \in \mathbb C : |z + 3 | = 3 \}$ is everything to the right-hand side of the $y$-axis, i.e. it is the region $\{ z \in \mathbb C : {\rm Re}(z) \geq 0 \}$.
A: No,
Mobius transform send circles to circles (Half plane is a special case of circles)
So The image is a half plane which is created by this points which is $x\ge0$

You can evaluate $f(1)$ to decide left side or right side

