How to see that image of $\frac{1}{z}$ under the set ${(x,y)\in \mathbb{C}; y = 2x+1}$ is a circunference? I need to see that 
$$f(z)=\frac{1}{z}$$
Under the set 
$${(x,y)\in \mathbb{C}; y = 2x+1}$$
Is a circunference. I started by doing:
$$f(x+i(2x+1)) = \frac{1}{x+i(2x+1)}\frac{x-i(2x+1)}{x-i(2x+1)}= \frac{x-i(2x+1)}{x^2+(2x+1)^2}$$
This has a real and inaginary part that doesn't look like a circle. How would you transform this equation to see that this is a circumference without knowing that it should be a circunference?
 A: I'll write $t$ instead of $x$, beacuse I will treat this variable as a parameter. The curve in $\Bbb R^2$ would be
$$(x(t),y(t))=\gamma(t)=\left(\frac t{t^2+(2t+1)^2}, -\frac{2t+1}{t^2+(2t+1)^2}\right)$$
This does not look a circle because the parametrization is not "standard". In fact, this is not a circle, but a circle with a point deleted (see the note below).
Since the implicit equation of a circle in $\Bbb R^2$ is like
$$x^2+y^2+Ax+By+C=0$$
let's compute $x^2+y^2$ to see if we can complete it.
$$x^2+y^2=\frac{t^2+(2t+1)^2}{[t^2+(2t+1)^2]^2}=\frac1{t^2+(2t+1)^2}$$
We see that
$$x^2+y^2+2x+y=0$$
which is indeed the equation of a circle.
Note: 
We have shown that every point in the image of $\gamma$ is in a circle. But not every point of the circle is in the image of $\gamma$. See what happens with $(0,0)$. Is every other point of the circle in the image of $\gamma$?
A: Here an "abstract" proof of it: $f(z)=1/z$ is trivially a Mobius transformation and therefore it maps generalized circles (that is lines or circles in the complex extended plane) to generalized circles. Call your set (which is a line on $\mathbb{C}$) $L$. By picking three suitable points in the image of $f$ it's easy to see that they are not colinear and thus $f(L\cup\{\infty_\mathbb{C}\})$ has to be a circle, so $f(L)$ is a circle missing the point $f(\infty_\mathbb{C})$.
This uses some basics of Mobius transformations, which might be too advanced if you're starting with complex analysis but I've posted this answer bacause it delegates calculations to the fact that Mobius transformations conserve generalized circles (in the extended complex plane), whose many avaliable proofs online sure can show you how to deal with your particular example more comprehensively.
