Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$. 
Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto the set $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$.

My attempt:
Let $z= x+iy$ then we have \begin{align}f(z) &= \frac{1}{z+2}\\ &=\frac{\bar{z}-2}{(z+2)(\bar{z}-2)} \\ &= \frac{(x-2)+i(-y)}{(x+2)^2 +y^2} \\ &= \frac{(x-2)}{(x+2)^2+y^2} +i(-y)\end{align}
Now let \begin{align}u(x,y)= \frac{(x-2)}{(x+2)^2 +y^2}~~ \text{and}~~v(x,y) = -y\end{align}
then we have $$f(z)= u(x,y) + iv(x,y)$$
Now the part that I am stuck with, is showing that the circle of radius 2 centered at the origin gets mapped onto the vertical line through $\frac{1}{4}$ on the real axis.
What I've tried is \begin{align}x^2 + y^2 &= 4\end{align}
Then, since we're interested in the real part, we now only look at \begin{align}u(x,y) &= \frac{(x-2)}{(x+2)^2 + y^2} \\ &= \frac{(x-2)}{x^2 + 4x +4 +y^2} \\ &= \frac{(x-2)}{4x + 8} \\ &= \frac{(x-2)}{4(x+2)}\end{align}
Now this is where I see I have a problem - I cannot cancel those two out, did I make a mistake somewhere?
 A: Depending on what theorems you have available, you might be able to use the following relatively easy (or at least low-computation) argument: Any fractional linear transformation, like the $f$ in your question, maps circles to circles or straight lines.  In the present case, one of the points on your circle $|z|=2$, namely the point $-2$, is mapped to $\infty$, so your circle gets mapped to a straight line. Another point on your circle, namely $2$, is mapped to $1/4$, so the image is a straight line through the point $1/4$.  At this stage, you can either compute the image of one more point on the circle and thereby uniquely determine the image line, or you can observe that $f$ clearly maps real numbers to real numbers, and, being analytic, preserves angles; since your circle is perpendicular to the real axis, its image is perpendicular to the image of the real axis, i.e., the image line is vertical.
A: Hint: It might be easier to parametrize the circle first, as $x=2\cos t, y=2 \sin t i.$ Then put this into $1/(z+2)$ and try to simplify the trig.
A: z+2 shifts the given circle two units to the right that passes through the origin . Now 1/(z+2) takes this circle to a line that passes through the point 1/4 and perpenicular to the x-axis.
A: I think you have an algebra error in your third line:
$$(z+2)(\bar{z}-2)=x^2+y^2-4iy-4\neq(x+2)^2+y^2.$$
As suggested, try using polar coordinates: the circle of radius 2 can be written $2e^{it}$ for $-\pi<t<\pi$. Plug this in to $f$ and take the real part:
$$Re\left(f(2e^{it})\right)=\frac{1}{2}Re\left(\frac{1}{e^{it}+1}\right)=\cdots=\frac{1}{4}.$$
You should be able to get that the whole line is filled in by considering $Im(f(2e^{it}))$.
