Division of two complex numbers as a consequence of the definition of multiplication.

Question

The definition of addition in $\mathbb{C}$ is: $$ (a,b)+(c,d)=(a+c,b+d) $$ When subtracting two numbers, the formula is quite straightforward from this definition: $$ (a+c,b+d)-(c,d)=((a+c)-c,(b+d)-c)=(a,b) $$ On the other hand, the definition of multiplication in $\mathbb{C}$ is: $$ (a,b)(c,d)=(ac-bd,ad+bc) $$ But the division of two numbers is found indirectly by multiplying the numerator by the complex conjugate of the denominator: $$ \frac{(a,b)}{(c,d)}=\frac{(a,b)(c,-d)}{c^2+d^2} $$ This way of finding the division confuses me: How do we know that multiply both the top and bottom by a complex number does not change the division? Sure this is the case in $\mathbb{R}$, but here the number is complex. This formula is invariantly introduced in even the most elementary complex variable textbooks without any justification for its validity.

My question is: What is the formal definition for the division of two complex numbers? By this I mean a formula that is a direct consequence of the definition of multiplication in $\mathbb{C}$.

Answer 1

Division is multiplication by the inverse, so in order to get a formula for division, you need a formula for the inverse.

Looking at the definition of multiplication, the inverse of $(c,d)$ is $(e,f)$ satsifying: $$ ec - fd =1$$ $$ ed + fc =0$$

Solving this for $e$ and $f$ gives that $(e,f) = \frac{(c,-d)}{c^2+d^2}$, which is a way to derive the formula you wrote without multiplying the numerator and denominator by the complex conjugate of $(c,d)$.

Answer 2

In principle, you could define the polar representation of complex numbers, $re^{i\theta}$, before defining division. The product of two complex numbers is $$r_1e^{i\theta_1}\times r_2e^{i\theta_2}=(r_1\times r_2)e^{i(\theta_1+\theta_2)}$$ and so the quotient is quite directly $$r_1e^{i\theta_1}\div r_2e^{i\theta_2}=(r_1\div r_2)e^{i(\theta_1-\theta_2)}$$ ...but I don't know if any textbooks actually take this approach.

Answer 3

What you need to understand is that $\mathbb{C}$ is a field, i.e. the triple \begin{equation} (\mathbb{C},+,\cdot)\end{equation} has to satisfy the following properties:

1. $(\mathbb{C},+)$ is an Abelian group with identity element $(0,0)$, where by definition \begin{equation} (a,b)+(c,d):=(a+b,c+d)\end{equation}
2. $(\mathbb{C}\setminus\lbrace(0,0)\rbrace,\cdot)$ is an Abelian group with identity element $(1,0)$, where by definition \begin{equation} (a,b)\cdot(c,d):=(ac-bd,ad+bc) \end{equation} the operations inside the parenthesis are those of $\mathbb{R}$.

A group is a couple $(G,\star)$ where $G$ is a set and the law of composition \begin{equation} \star\colon G\times G\to G \end{equation} has to satisfy the following properties

1. $\star$ is associative: $\forall g_1,g_2,g_3\in G\quad (g_1\star g_2)\star g_3=g_1\star(g_2\star g_3)$;
2. $\star$ has an identity element $e$ such that $\forall g\in G,\quad g\star e=e\star g=g$;
3. existence of an inverse element: $\forall g\in G,\,\exists g^{-1}\in G: g\star g^{-1}=g^{-1}\star g=e$.

If moreover $\forall g_1,g_2\in G,\quad g_1\star g_2=g_2\star g_1$, the group is called Abelian.

Observe that $(0,0)$ is the only element in $\mathbb{C}$ which has no multiplicative inverse. The operation of division is just an alias for multiplication of an element by the inverse of another, it doesn't add more structure to $\mathbb{C}$.

So, as explained by Y. Forman, you can define \begin{equation} (a,b)/(c,d):=(a,b)\cdot (c,d)^{-1} \end{equation} then you are looking for an expression of $(c,d)^{-1}=(x,y)$ in terms of the real numbers $c,d$: you have just to impose \begin{align} (c,d)\cdot(x,y)&=(1,0)\\ &=(cx-dy,cy+dx)\end{align} therefore \begin{align} cx-dy&=1\\ cy+dx&=0 \end{align} which yields $x=c/(c^2+d^2)$ and $y=-d/(c^2+d^2)$, i.e. \begin{equation} (c,d)^{-1}=\frac{1}{(c,d)}=(\frac{c}{c^2+d^2},-\frac{d}{c^2+d^2}), \end{equation} and \begin{equation} \frac{(a,b)}{(c,d)}=(\frac{ac+bd}{c^2+d^2},\frac{-ad+bc}{c^2+d^2})=\frac{(a,b)\cdot(c,-d)}{c^2+d^2}. \end{equation}

Answer 4

I have been looking for this proof for a long time since it did not make sense to me too but now I see why we are multiplying by the conjugate. The complex numbers follow some properties of real numbers and the inverse holds true. Thank you all for the info