<s>[INST] 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?
 [/INST] 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.
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