# Algebraic Method for Inverse Maps

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