Application of z In which lines of the $w$ plane is the line $|z| = \frac{1}{2} $ transformed with the function $w= \frac{1}{z} $?
$$| z | = \frac{1}{2}$$
$$\sqrt{x^2 + y ^2} = \frac{1}{2}$$
$$x = \sqrt{\frac{1}{4} - y^2}$$
$$y = \sqrt{\frac{1}{4} - x^2}$$

$$w = u+iv = \frac{1}{z}=\frac{1}{u+iv}=\frac{x-iy}{x^2+y^2}$$
$$u=\frac{x}{x^2+y^2}$$
$$v=-\frac{y}{x^2+y^2}$$
such that
$$w = \frac{\sqrt{\frac{1}{4}-y^2}}{\frac{1}{2}-x^2-y^2} -i \frac{\sqrt{\frac{1}{4}-x^2}}{\frac{1}{2}-x^2-y^2}$$
However, I don't see how $w = u^2 + v^2 = 4$, which is a circle and the answer according to the problem set.

$$z=\frac 1w\Rightarrow \left|\frac 1w\right|=\frac 12\Rightarrow |w|=2\Rightarrow u^2+v^2=4$$
