# Image of unit disk under inversion - Mobius transformations

I am trying to find the image of $$\{z \in \mathbb{C} : |z| < 1\}$$ under $$f(z) = \frac{1}{z}$$

Let $$w = \frac{1}{z} \Rightarrow z = \frac{1}{w} = \frac{1}{u+iv} = \frac{u-iv}{u^2 + v^2}$$

if $$w = u +iv$$.

Then considering the boundary first, $$|z| =1 \Rightarrow \Big|\frac{u}{u^2+v^2} -i\frac{v}{u^2+v^2}\Big| =1$$ Using Pythagoras gives $$\left(\frac{u}{u^2 +v^2}\right)^2 + \left(\frac{v}{u^2 +v^2}\right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $$(u^2 + v^2)^2$$ lying about, which is what the answer should be...

What's happening?

Of course, it makes no sense to talk about the image of $$0$$ under $$\frac1z$$, but the image of $$\{z\in\mathbb{C}\,|\,\lvert z\rvert<1\}\setminus\{0\}$$ is $$\{z\in\mathbb{C}\,|\,\lvert z\rvert>1\}$$.