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I just started a book on Complex Variables and there is something I cannot grasp:

The product of $z_1z_2$ is defined as follows: $z_1 = (x_1,y_1)$ and $z_2 = (x_2,y_2)$

Now how did they manage to get this: $(x_1,y_1)(x_2,y_2)=(x_1x_2−y_1y_2, y_1x_2+x_1y_2)$.

Where did the minus even come from when doing the product? Can someone please explain?

José Carlos Santos
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Mac
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    This is how the product of two complex numbers is defined. – Math Lover Sep 06 '17 at 22:13
  • @MathLover Exactly and I am having trouble wrapping my head around the definition because this is as rudimentary as the textbook gets – Mac Sep 06 '17 at 22:16
  • This (Hamilton pair) representation is a special case of canonical remainder reps when computing modulo polynomials, here we are computing with polynomials modulo $,\color{#c00}{x^2+1},$ where $\color{#c00}x$ represents $\color{#c00}i.,$. See this answer for a detailed explanation. – Bill Dubuque Sep 06 '17 at 23:14
  • $\rm\begin{eqnarray}\rm (a,\ b) &&\rm (c,\ d) &!!=&\rm (ac!\color{#000}{\bf -}!bd,\quad\ \ ad!+!bc)\ \rm i.e.\ \bmod{\color{#c00}{,x^2!+1}}!:\ \ \color{#c00}{x^2\equiv -1}\ ,\Rightarrow,\ (a! +! b\color{#c00}x)&&\rm(c! +! d\color{#c00}{ x}), &!!!\rm,\equiv&\rm (ac!\color{#c00}{\bf -}!bd) + (ad!+!bc), x\
    \rm i.e.\quad, (a! +! b{\it\color{#c00} i},)&&\rm(c! +! d {\it\color{#c00} i},), &!! =,&\rm (ac!\color{#000}{\bf -}!bd) + (ad!+!bc),{\it\color{#c00} i}\end{eqnarray}\ \ $     – Bill Dubuque Sep 06 '17 at 23:15
  • See also this answer for the interesting historical background and motivation for Hamilton's pair construction of $,\Bbb C.\ \ $ – Bill Dubuque Sep 06 '17 at 23:23

2 Answers2

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$$(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i$$

So $(a,b)$ and $(c,d)$ are transformed into $(ac-bd,ad+bc)$

Peter
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You identify $(x,y)$ with $x+iy$, where $i$ have the property that $i^2=-1$. Then $$(x,y)\cdot(z,w)=(x+iy)\cdot(z+iw)=xz+ixw+iyz+i^2yw=(xz-yw)+i(xw+yz)=(xz-yw,xw+yz)$$

José Alejandro Aburto Araneda
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