Find and sketch the line $x=1$ under the mapping $f(z)=1/z$ Find and sketch the image of the vertical line $x=1$ under the mapping $f(z)=\frac 1z$
I started by using $u(x,y) + iv(x,y)=f(z)=\frac 1z = \frac 1{x+iy}$ From here I multiplied $f(z)$ by the conjugate to get $\frac x {x^2 + y^2} - \frac {iy} {x^2 + y^2}$ Then I took $u(x,y)=\frac {x} {x^2 + y^2}$ and $v(x,y)=- \frac {iy} {x^2 + y^2}$ Now, plugging in $x=1$ for $U(1,y)$ and $v(1,y)$ to get $u(1,y)=\frac {1} {1 + y^2}$n and $v(1,y)=\frac {-y} {1 + y^2}$ and then Im stuck. Im not sure how to go from here to a graph.
 A: Using $u_y=u(1,y)$ and $v_y=v(1,y)$, you obtain the equation
$(u_y-\frac{1}{2})^{2} +v_y^{2} =(\frac{1}{2})^{2}$ for any $y\in\mathbb{R}$.
Conversely you have for any point $z=x+iy$ in the circle with center $(\frac{1}{2},0)$ and radius $\frac{1}{2}$
\begin{equation}
(x-\frac{1}{2})^{2} +y^{2} =(\frac{1}{2})^{2}\\
\Rightarrow x^2-x+y^2=0
\end{equation}
so
$\frac{1}{z}=\frac{x-iy}{x^2+y^2}=1-\frac{iy}{x^2+y^2}$.
Therefore $f(z)=\frac{1}{z}$ is a bijection of the vertical line $x=1$ and the circle $(x-\frac{1}{2})^2 +y^2=(\frac{1}{2})^2$.
A: One way to deduce the circle in Pedro's answer is to be already be aware that any Möbius transform, of which the inversion $f(z)=1/z$ is an example, maps lines and circles to lines and circles (not necessarily respectively, i.e. a circle could be mapped to a line or circle.) So it's enough to map a finite number of points to find out which line/circle is created. Since
$$f(1)=1,\quad f(1\pm i)=\frac{1}{2}\mp \frac{1}{2}i, \quad f(\infty)=0,$$
we conclude that the image of $x=1$ is a circle passing through the points $(1,0),(\frac{1}{2},\pm \frac{1}{2}),(0,0)$. From this we may deduce the same circle as given in Pedro's answer.
A: So $(1+iy)\mapsto \frac{1}{1+y^2}+i\frac{-y}{1+y^2}$.
Now here's a hint: What is the absolute value of this image point?
