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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?

John P.
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    “In $\mathbb{C}$, division is a bit less clear-cut.” – Is it? Why do you think so? – Martin R Mar 22 '20 at 21:02
  • It seems to me that there are additional complications. Is it as simple as $z_2$ and $z_4$ being invertible (non-zero)? – John P. Mar 22 '20 at 21:03
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    Why would it not be that simple? – PrincessEev Mar 22 '20 at 21:06
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    $\Bbb C$ is a field, and the left-hand side is not defined if $z_2$ or $z_4$ is zero. Or did you think of the extended complex plane/Riemann sphere, where (simplifying things a bit) something like $1/0 = \infty$ can make sense? – Martin R Mar 22 '20 at 21:07
  • I wasn'y thinking of the extneded complex plane, but did want to define these fractions so that neither $z_2$ nor $z_4$ were zero. Does it work in that case? – John P. Mar 22 '20 at 21:27
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    This is true in any field and I will upvote only answers which only use the field axioms. – Torsten Schoeneberg Mar 22 '20 at 21:40

2 Answers2

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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.

Torsten Schoeneberg
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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 :))

Bumblebee
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