Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$ 
Find the image of $|z+1|=2$ under $f(z) = \frac{1}{z}$ where $z \in \mathbb C$

My attempt:
Let $z = x + iy$
$\displaystyle |z+1|=2 \iff | (x + iy)+1|=2 \iff |(x+1) +iy|=2 \iff (x+1)^2 + y^2 = 4$
Let $w = u + iv$
Now let $\displaystyle w = \frac{1}{z}$ hence we have that \begin{align}z &= \frac{1}{w} \\ &= \frac{1}{u + iv} \\ &= \frac{u-iv}{u^2 + v^2} \\ &= \frac{u}{u^2 + v^2} + i \big( - \frac{v}{u^2 + v^2} \big)\end{align}
From which we can deduce that $\displaystyle x = \frac{u}{u^2 + v^2}$ and $\displaystyle y = - \frac{v}{u^2 + v^2}$
and thus $$\displaystyle \bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4$$
This is where I am stuck. I keep on messing up the simplification. Can someone please show me how to simplify this?
 A: Proceeding with your result, 
$$
 \bigg(\frac{u}{u^2 + v^2} +1\bigg)^2 + \bigg(- \frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ we get
$$
 \bigg(\frac{u}{u^2 + v^2}\bigg)^2+2\frac{u}{u^2 + v^2}+1 + \bigg(\frac{v}{u^2 + v^2}\bigg)^2 = 4
$$ or
$$
\frac{u^2}{(u^2 + v^2)^2}+ \frac{v^2}{(u^2 + v^2)^2}+\frac{2u}{u^2 + v^2}  = 3
\\$$$$
2u+1 = 3u^2+3v^2
$$ or, with a little algebra,

$$
\left(u-\frac13 \right)^2+v^2=\frac49
$$ 

which is an easy set to identify.
A: Another approach, using inversive geometry: Observe that $f(z)$ almost produces the inversion of a figure, with respect to the unit circle; in fact, it produces the complex conjugate of the figure's inversion.  However, in our case, the circle $|z+1| = 2$ is already symmetric with respect to the real axis, so $f(z)$ will in fact produce the figure's inversion.
Now, note that our input circle is tangent to the unit circle at $z = 1$, so the input circle's inversion, our output circle, will also be tangent to the unit circle at $z = 1$, but "on the inside."  Since the input circle also intersects the real axis at $z = -3$, the output circle will intersect the real axis at $z = -1/3$, so the equation for the image must be
$$
\left|\, z-\frac{1}{3} \right| = \frac{2}{3}
$$
