On division by $0$ in $\hat{\mathbb{C}}$

Question

I've recently sat through an introductory course in Complex Analysis, and upon introducing Möbius Transformations, I was made aware of the set $\hat{\mathbb{C}}$. The set $\hat{\mathbb{C}}$ was defined to me as $\hat{\mathbb{C}}=\mathbb{C}\cup\lbrace\infty\rbrace$, to accompany for the limit as $z$ tends to infinity of a Möbius Transformation. Another notion that was introduced was the notion of division by zero. In one of the questions, we had the following

If $$f(z)=\frac{az+b}{cz+d}$$and $f(i)=\infty$, what does this imply about the values $a, b, c,$ and $d$?

And the correct response was that $d=-ci$, given that this would yield a denominator of $0$, allowing for $f(i)$ to equal $\infty$. Now, I understand that when we introduce infinity as an element of a set, conventional ideas would break down a tad, but this seemed rather extreme to me. I'm not yet able to understand the bare bones of features of math such as the Riemann Sphere that allow for expressions such as this to be "well-behaved," but can somebody perhaps briefly explain to me why this is possible?

Any responses are appreciated, thank you.

Answer 1

In $\hat{\Bbb C}$ we simply define $$\frac z0=\infty\quad(z\ne0).$$

How is this possible? We can make any definition we like, so there.

The real question is how this is possible without leading to a contradiction, given that we can prove that division by zero is impossible.

A proof that division by zero is impossible is really a proof that if we assume various axioms then division by zero is impossible. For example, if we assume that $(a/b)b=a$ for every $a$ and $b$ then it's impossible to define division by zero. But in $\hat{\Bbb C}$ the axiom $(a/b)b=a$ simply does not hold; that's the price we pay for making this definition. What we gain is that the formalism of Mobius transformations works out nicer; we can actually define $f(z)=(az+b)/(cz+d)$ for every $z$, no need to special-case mumble about what this means if the denominator vanishes...

Answer 2

The Riemann sphere (=complex projective line) is not a field, so you are not really "dividing by zero", which is clearly strictly forbidden in every such a structure.

What you are saying is simply that the (affine) Moebius transformation $f \colon \mathbb C \longrightarrow \mathbb C$ extends to a bijective mapping $\bar{f} \colon \mathbb{P}^1(\mathbb C) \longrightarrow \mathbb{P}^1(\mathbb C)$.

Note that $\infty$ is not a special point in $\mathbb{P}^1(\mathbb C)$, in the sense that you can make a change of coordinates and make it any "finite" point: for instance, the change of coordinates $z \mapsto 1/z$ sends $\infty$ to $0$. Using the homogeneous coordinates notation, this means that it sends $[1:0]$ to $[0:1]$.