Finding the image of the unit disc under a Mobius transform

I am working through an example of mapping the unit disc under the following Mobius transform: $$f(z)=\frac{iz+3}{iz-1}=1+\frac{4}{iz-1}$$ This can be written as a composition of elementary Mobius transforms: $$f_5 ∘f_4 ∘ f_3 ∘ f_2 ∘ f_1$$ where $$f_1(z)=iz, f_2(z)=z-1, f_3(z)=1/z, f_4(z)=4z, f_5(z)=1+z$$.

Working through each of the mappings, $$f_1$$ maps the unit disc onto the unit disc. $$f_2$$ shifts the unit disc to the left by 1, leaving us with a unit disc centred at $$z=-1$$.

This is where I get confused - the mapping $$f_3$$ supposedly maps the unit disc onto the half space where $$Re(z)<-1/2$$, and I am struggling to see how. Indeed, the image of -2 will be -1/2. But how do we know it does not map to another circle, and how do we know it maps to $$Re(z)<-1/2$$ and not $$Re(z)>-1/2$$?

I can sort of convince myself that it will be a plane and not a circle, due to the fact that the circle has the point $$z=0$$ which upon the inverse mapping will go to $$z=∞$$, but I find this to be a weak argument, and neither can I seem to convince myself it will be the left of $$z=-1/2$$.

I would greatly appreciate any insight on this.

Thank you.

• Consider the mapping of the boundaries first. The image of $|z + 1| = 1$ is either a straight line or a (regular) circle, and it cannot be a circle, because circles do not go through infinity. To determine the half-plane, take the image of an arbitrary point in the interior (or consider that the circle is locally rotated by $\pi$ at $z = -2$). – Maxim May 1 at 19:03