Equality of fractions in $\mathbb{C}$

Question

In $\mathbb{R}$, $$\frac{a}{b} = \frac{c}{d} \iff ad = bc.$$ In $\mathbb{C}$, division is a bit less clear-cut. Nonetheless, if we have complex numbers $z_1, \ldots, z_4$, can we still assert that $$\frac{z_1}{z_2} = \frac{z_3}{z_4} \iff z_1 z_4 = z_2 z_3 $$ provided that $z_2, z_4 \neq 0$? Are there any counterexamples to this?

Answer 1

One way to define complex numbers is, equip $\mathbb{R}^2$ with coordinate addition and multiplication given by $$(a,b)(c,d)=(ac-bd, ad+bc).$$ (Can you see the hidden trick here :)) This defines a nice commutative field and, our good old friend, real numbers $\mathbb{R}$ sits inside this new system in the form of pairs $(a,0).$ One can manually verify that $$(a,b)\left(\dfrac{a}{a^2+b^2},\dfrac{-b}{a^2+b^2}\right)=(1,0),\qquad a^2+b^2\neq0$$ and hence the second ordered pair defines the multiplicative inverse of the first. Now that we have multiplicative inverses for non-zero numbers we can think of division as multiplying by the multiplicative inverse. If you haven't convince by this interpretation odf division of complex numbers, there are other ways to think of complex numbers as "a sub-ring of real $2\times 2$ matrices" or a "quotient ring of real polynomials" and define division properly inside these rings.

Later: Since you are wondering about, equivalent fractions you can verify by hand that $$\dfrac{z_1}{z_2}=\dfrac{z_3}{z_4}\iff z_1z_4=z_2z_3,\qquad z_2,z_4\neq 0$$ using this ordered pair interpretation (but will be tedious as you need to manipulate $8$ real numbers :))

Answer 2

Just multiply both sides with $z_2 z_4$ and use that this multiplication is bijective provided neither of $z_2,z_4$ is $=0$. The same proof works in any commutative ring where the denominators are invertible, i.e. the fractions make sense to begin with; in particular, any field.