How to draw the image of a Möbius transformation? Let $f(z)=\frac{2z}{z+2}$. What is the image of the circle with radius 2 centered at 0?

I find that $f$ maps $2\to1$, $-2 \to \infty$. Hence, the image should be a line. Therefore, I pick another point on the circle, say $2i$, and find $f(2i)$. Then the line through $1$ and $f(2i)$ is the image?
Your function goes from the z in the w plane, so we need to find z to substitute in the circle equation.
$$w = \frac{2z}{z+2} \Rightarrow z = \frac{2w}{2-w}$$
The equation for the circle is $|z|=2 \Rightarrow |z|^2=4 \Rightarrow z\overline{z} = 4$. Substitute the inverse function:
$$\left(\frac{2w}{2-w}\right)\overline{\left(\frac{2w}{2-w}\right)} = 4$$
$$\left(\frac{2w}{2-w}\right)\left(\frac{2\overline{w}}{2-\overline{w}}\right) = 4$$
$$4w\overline{w} = 16 -8\overline{w} -8w + 4w\overline{w}$$
$$8w+8\overline{w} = 16$$
then let $w = u + vi$:
$$8(u+vi)+8(u-vi)=16$$
And the result is $u=1$

$f(2) = 1$ and $f(2i) = 1 + i$ so you get the same line.
First we know that this Möbius transformation maps the x-axis to x-axis. Then we know two points on the circle, 2 and -2, are mapped onto 1 and $\infty$. Since the Möbius transformation preserves angles, the image of the circle should intersect with the x-axis by 90 degrees. Also, the Möbius transformation maps circles to lines or circles. Since there is $\infty$ in the image, the image couldn't be a circle. Hence, the image should be a straight line passing through 1 and parallel to the y-axis. That is, $w=1+vi,v\in \mathbb{R}$ plus $\infty$.