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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
\includegraphics[width=5in]{../../../texdocs/ip9}


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

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

Then,

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

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

(i)$\displaystyle \quad$ $\displaystyle z_1+z_2$   and$\displaystyle \quad z_1z_2\in s$   (multiplicative and additive closure)    
(ii)$\displaystyle \quad$ $\displaystyle z_1+z_2=z_2+z_1$   and$\displaystyle \quad z_1z_2=z_2z_1$   (commutativity)    
(iii)$\displaystyle \quad$ $\displaystyle z_1+(z_2+z_3)=(z_1+z_2)+z_3$   and $\displaystyle z_1(z_2z_3)=(z_1z_2)z_3$   (associativity)    
(iv)$\displaystyle \quad$ $\displaystyle z_1(z_2+z_3)=z_1z_2+z_1z_3$   (distributive)    
(v)$\displaystyle \quad$ $\displaystyle (0,0)  $ is the additive identity.    
(vi)$\displaystyle \quad$ $\displaystyle (a,b)\in S$ then $\displaystyle (-a,-b)\in S$ is its additive inverse.    
(vii)$\displaystyle \quad$ $\displaystyle (1,0)$ is the multiplicative identity.    
(viii)$\displaystyle \quad$ $\displaystyle (a,b)\in S,   a$ and $\displaystyle b$ not both$\displaystyle    0,$    then the multiplicative inverse is    
  $\displaystyle \colorbox{Goldenrod}{\boxed{\left(\frac{a-2b\cos\theta}{a^2+b^2-2ab\cos\theta},\frac{-b} {a^2+b^2-2ab\cos\theta}\right).}}$    

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

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

Hence,

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

So, from Cramer's rule,

$\displaystyle 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}.
$

$\displaystyle 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,

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

Hence

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

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

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

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

$\displaystyle \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

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

and the imaginary part of $ z$ is

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


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

$\displaystyle Re(z)=\frac{(-\bar{h})z+h\bar{z}}{h-\bar{h}}=\frac{z+\bar{z}}{2}$   and$\displaystyle \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}$
\includegraphics[width=4.5in]{../../../texdocs/ip10}

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
\includegraphics[width=3.0in]{../../../texdocs/ip11}

In figure (6)

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

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

Then

$\displaystyle 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

$\displaystyle 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)$
\includegraphics[width=3.5in]{../../../texdocs/ip12}

next up previous contents
Next: The difference of two Up: Pythagorean Triples, etc. Previous: 120 degree triples and   Contents
f. barnes 2008-04-29
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