Find the image of $x+y=4$ under the mapping $w=z^{-1}$.
I have read of an algebraic method that seems to work (unsure if my answer is correct).
Let $$x+iy=z=\frac{1}{w}=\frac{1}{u+iv}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$$ Now upon equating real and imaginary components, $$x=\frac{u}{u^2+v^2}, \ \ y=-\frac{v}{u^2+v^2}$$
Now, \begin{align} x+y&=4 \\ \frac{u}{u^2+v^2}-\frac{v}{u^2+v^2}&=4 \\ u-v&=4(u^2+v^2) \\ 4u^2-u+4v^2+v&=0 \\ \left(u-\frac{1}{8}\right)^2+\left(v+\frac{1}{8}\right)^2&=\frac{1}{32} \ \ \ \ \ \ \ \text{(upon completing the square)} \end{align}
Is the method valid? It is very simple. Any advice would be really appreciated
The posted proof looks good.
For an alternative algebraic solution without resorting to cartesian:
write the condition $\,x+y=4\,$ as $\,z+\bar z - i(z - \bar z) = 8\,$;
substitute $\,z = 1/w\,$ and simplify to $\,8 w \bar w - (1+i)w - (1-i) \bar w = 0\,$;
rearrange the above to write as $\,\left|8w - (1-i)\right|^2 = 2\,$.