How to teach newbie multiply of complex number

Question

I want to teach a newbie the arithmetic law of complex numbers. the law of add is acceptable psychological. but multiply is not. for example, assume $$z = a+bi, w = c+di$$

He (She) may ask me: why not $$zw = ac+bdi$$, why must we define multiply like this: $$zw = ac-bd +i(ad+bc)$$. It is hard to acceptable.

I want to explain, but I am failed. Please help me, thanks very much.

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My explain:

First , define like this,then the multiply of complex number can satisfy most laws of algebra for addition, multiplication. Second, If I treat complex number as a linear operation of $\Bbb{R}^2$. then the multiplication of complex number is coincide with the multiplication of matrix in $M_2(\Bbb{R})$. this explain that this definition is reasonable.

Answer 1

I do not know how much this could help but I faced the same problem.

Let say that we have $z = a+bi, w = c+di$ and that we want to compute the product $z \times w$. So we have $$z \times w=(a+ bi) \times (c+di)=a \times (c+di)+bi \times (c+di)$$ Now, continue expanding and later collecting the terms.

Answer 2

\begin{align} (a+i*b)*(c+i*d)&=ac + i*ad + i*bc + i*b*i*d \\ & = ac + i(ad+bc) + i*i*bd \\ & = ac + i(ad+bc) + i^2*bd \\ & = ac - bd + i(ad+bc) \end{align} The first step is using the law of distributivity, the last step is using $i^2=-1$

Answer 3

I suggest drawing a parallel between $\mathbb C = \mathbb R [i]$ and, say $\mathbb Q[\sqrt 2] = \{a+b\sqrt 2| a,b \in \mathbb Q\}$. They are both sets made by extending a set by one "weird" number and in both cases the square of this element is again a "normal" number. The two sets share many properties.

However, because $\mathbb Q[\sqrt 2] \subset \mathbb R$, this set should be simpler to grasp. You can show that if we were to define $(a+b\sqrt 2) \times (c+d\sqrt 2) = ac + db \sqrt 2$, this would not be the multiplication on $\mathbb R$.

Answer 4

You yourself appear to have a couple of misconceptions about the problem!

You're not defining anything. Well, except for $i^2=-1$. You're simply calculating the product:

$$(a+bi)(c+di)$$

By distributing and then gathering terms, see Genju's answer.

Your definition of addition isn't a definition either. It's just distributing and commuting.

$$a+ib+c+id=(a+c)+i(b+d)$$

(Of course, strictly speaking it's not very rigorous to construct $\mathbb C$ by just "adjoining" $i$ and saying "$i^2=-1$ but otherwise everything works as normal", but that's how $\mathbb C$ is often constructed when students are seeing complex numbers for the first time.)

Answer 5

The way that Complex numbers are generated is to take the Field of Real numbers, and add a point ($i=\sqrt{-1}$).

Formally, a field is a set with 2 operations defined on its elements (+,$\times$). The elements have the following properties $$\begin{array}{rrr} commutative&a\times b=b\times a & a+b = b+a \\ associative&(a\times b)\times c=a\times (b\times c) & (a+b)+c = (a+b)+c \\ identity&1\times a = a & 0+a=a \\ inverses&a(a^{-1}) = 1 (a\ne 0) & a+(-a)=0 \\ \end{array}$$ and $$distributive\:\:\:a\times(b+c) = (a\times b)+(a\times c)$$ When adding the point $i$ we need to assume that the preceding axioms hold (so long as the new field is consistent [i.e. doesn't contradict itself]).

This means that if $i$ is a number, then so is $bi$. If $bi$ is a number, then so is $a+bi$ (where $a,b \in \mathbb R$).

So now that we know that $\mathbb C$ is a Field (whose elements obey Field axioms), we can turn our attention to the problem. $$z=a+bi,w=c+di$$

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