Image of Upper Half Disc under $w = 1/z$ I need to find the image of the upper half disc $|z|<1$, $Im\, z >0$ under the inverse transformation $w = 1/z$.
Now, since $|z|<1$, $|z|^{2}<1$. Rewriting this as $z\overline{z}<1$, we make the substitution $z = 1/w$ to obtain $\frac{1}{u^{2}+v^{2}}<1$, or $1 < u^{2}+v^{2}$, which is the exterior of the unit circle in the $w$-plane.
Also, we need to have $Im\, z = \frac{z-\overline{z}}{2i} > 0$. Again, making the substitution $z = 1/w$ and simplifying, we get that $-\frac{v}{u^{2} + v^{2}} > 0$, which implies that $\frac{v}{u^{2}+v^{2}} < 0$, and since $u^{2} + v^{2}$ is always positive, that $v < 0$.
So, what I think is that the image of the upper half disc under $w = 1/z$ is the exterior of the lower half disc $|w|>1$, $Im\, w < 0$.  Could someone please tell me if I'm right? And if not, where did I go wrong?
Thank you.

Your answer is true but you could go faster with polar form.
If $z = re^{i\theta}$ with $r<1$ and $\theta \in ]0,\pi[$ then $\frac{1}{z} = \frac{1}{r} e^{-i\theta}$ where $\frac{1}{r} > 1$ and $-\theta \in ]-\pi,0[.$
