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.
Best Answer
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 $(x-\frac{1}{2})^2 +y^2=(\frac{1}{2})^2$.