The complex numbers from the law of cosines

This section in pdf form. complex.pdf

It's shown that generalized complex numbers arise naturally from the law of cosines, $x^2+y^2-2ab\cos\theta=z^2$ , where side $z$ is opposite $\angle\theta$ . These numbers can be plotted on an "extended" Argand diagram using the natural coordinates where the imaginary axis is rotated clockwise $\theta$ degrees out of the negative real axis. Similarly, the conjugate axis is rotated counter clockwise $\theta$ degrees out of the negative real axis. Hence if $\angle \theta=90 ^{\circ}$ then the imaginary and conjugate axes are coincident resulting in the "standard" Argand diagram, and the "standard" complex numbers.

Figure 4: two triangles

Consider the triangles in fig (4). From the law of cosines, $z^2=a^2+b^2-2ab\cos\theta$ and $w^2=c^2+d^2-2cd\cos\theta$ . If

$$x=ac-bd y=ad+bc-2bd\cos\theta.$$ (42)

Then,

$$x^2+y^2-2xy\cos\theta=\left(a^2+b^2-2ab\cos\theta\right)\left(c^2+d^2-2cd\cos\theta\right)=(zw)^2.$$ (43)

Let $S$ be the set of all ordered pairs (a,b) such that a and b are real numbers, and $(a,b)=(c,d)$ iff $a=b$ , and $c=d$ .

• Addition in $S$ is defined as $(a,b)+(c,d)=(a+c,b+d)$ .
• 1. For multiplication, I will use equation (42), that is, for $(a,b)$ and $(c,d)$ in $S$ , $\colorbox{Goldenrod}{\boxed{(a,b)(c,d)=(ac-bd,ad+bc-2bd\cos\theta)}}$ .
  2. $m(a,b)=(ma,mb), m\in \mathbb{R}$ .
  .

If $z_1$ , $z_2$ , and $z_3$ are elements of $S$ , then

$$\quad z_1+z_2 \quad z_1z_2\in s$$ (multiplicative and additive closure)

$$\quad z_1+z_2=z_2+z_1 \quad z_1z_2=z_2z_1$$ (commutativity)

$$\quad z_1+(z_2+z_3)=(z_1+z_2)+z_3 z_1(z_2z_3)=(z_1z_2)z_3$$ (associativity)

$$\quad z_1(z_2+z_3)=z_1z_2+z_1z_3$$ (distributive)

$$\quad (0,0)$$

$$\quad (a,b)\in S (-a,-b)\in S$$

$$\quad (1,0)$$

$$\quad (a,b)\in S, a b 0,$$

$$\colorbox{Goldenrod}{\boxed{\left(\frac{a-2b\cos\theta}{a^2+b^2-2ab\cos\theta},\frac{-b} {a^2+b^2-2ab\cos\theta}\right).}}$$

Hence $S$ , under addition and multiplication as defined, is a field.

The multiplicative inverse was found by noting that if $(c,d)$ is the multiplicative inverse of $(a,b)$ then

$$(a,b)(c,d)=(ac-bd,ad+bc-2bd\cos\theta)=(1,0).$$

Hence,

$$\begin{pmatrix} ac&-bd\\ bc&(a-2b\cos\theta)d \end{pmatrix}= \begin{pmatrix} 1\\ 0 \end{pmatrix}.$$

So, from Cramer's rule,

$$c=\frac{\begin{vmatrix} 1&-b\\ 0&(a-2b\cos\theta) \end{vmatrix}}... ...&(a-2b\cos\theta) \end{vmatrix}}=\frac{a-2b\cos\theta}{a^2+b^2-2ab\cos\theta}.$$

$$d=\frac{\begin{vmatrix} a&1\\ b&0 \end{vmatrix}}{\begin{vmatrix} a&-b\\ b&(a-2b\cos\theta) \end{vmatrix}}=\frac{-b}{a^2+b^2-2ab\cos\theta}.$$

$(a,b)$ can be written as a sum. Note that $(a,b)=a(1,0)+b(0,1)$ and $(1,0)$ is equivalent to $1$ ; so let $h=(0,1)$ then $(a,b)$ can be written as $a+bh$ . I want to find an expression for $h$ . We have,

$$h^2=(0,1)(0,1)=(-1,-2\cos\theta)=-1-(2\cos\theta) h$$

Hence

$$h=\frac{-2\cos\theta\pm \sqrt{4\cos^2\theta-4}}{2}=-\cos\theta\pm i\sin\theta$$ (44)

where $i=\sqrt{-1}$ . Set

$$\boxed{h=-\cos\theta+i\sin\theta\quad \mbox{and the conjugate}\quad \bar{h}=-\cos\theta-i\sin\theta}.$$

If $z=a+bh$ and $\bar{z}=a+b\bar{h}$ then

$$\sqrt{z\bar{z}}=\vert z\vert=\sqrt{a^2+b^2-2ab\cos\theta} .$$

And,From equation (43), $\vert z_1z_2\vert=\vert z_1\vert\vert z_2\vert$ .

The real part of $z$ is

$$a=Re(z)=\frac{(-\bar{h})z+h\bar{z}}{h-\bar{h}} ,$$

and the imaginary part of $z$ is

$$b=Im(z)=\frac{z-\bar{z}}{h-\bar{h}} .$$

Note: If $h=i$ and $\bar{h}=-i$ then

$$Re(z)=\frac{(-\bar{h})z+h\bar{z}}{h-\bar{h}}=\frac{z+\bar{z}}{2}$$   and$$\quad Im(z)= \frac{z-\bar{z}}{h-\bar{h}}=\frac{z-\bar{z}}{2i}.$$

Plotting

Figure 5: Plotting $z=a+bh$ and $\bar{z}=a+b\bar{h}$

From equation(43), we know that if $z=a+bh$ is plotted in the plane, then $\vert z\vert$ , the distance from the origin to the point $(a,b)$ , is $\sqrt{a^2+b^2-2ab\cos\theta}$ . This can be accomplished if the positive $yh$ axis is rotated $\theta$ degrees clockwise out of the $-x$ axis as in figure (5). Rotate the positive conjugate axis counterclockwise 90 degrees out of the negative real axis. Then $\bar{z}$ can be plotted along with $z$ , as shown in figure (5).

Polar coordinates

Figure 6: Finding a+bh in polar coordinates

In figure (6)

$$b=\frac{q}{\sin\theta}=\frac{r\sin\beta}{\sin\theta}$$    where $$\sin\theta\neq 0.$$

$$a=r\cos\beta+b\cos\theta=r\left(\cos\beta+\sin\beta\cot\theta\right).$$

Then

$$a+bh=r\left(\cos\beta+\sin\beta\cot\theta+\frac{\sin\beta}{\sin\theta} (-\cos\theta+i\sin\theta)\right)=r(\cos\beta+i\sin\beta).$$

And

$$a+b\bar{h}=r(\cos\beta-i\sin\beta).$$

Conclusion

In equation (44), for $0 \geq\theta<\pi$ , $h=i$ if and only if $\theta=\pi/2$ . That is, if and only if the imaginary and conjugate axes are rotated $90$ degrees from the $-x$ axis, clockwise and counterclockwise respectively in figure (5). This brings them into coincidence at right angles to the real axis as shown in figure (7).

Replace $\theta$ with $90^{ \circ}$ throughout. Since $\cos(90^{ \circ})=0$ , multiplication becomes $(a,b)(c,d)=(ac-bd,ad+bc)$ , $h=i$ , and $\bar{h}=-i$ . Then $S$ becomes the field of complex numbers.

Figure 7: Plotting $z=a+bi$ and $\bar{z}=a+b(-i)$

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