#129
\mathbb{C}_3\circ\R
Base: ij = i, ji = j
\begin{aligned}
z &= (a, b, c)\cdot(1, i, j) &= a + bi + cj\\
w &= (d, e, f)\cdot(1, i, j) &= d + ei + bj\\
\bar{z} &= (a, b, c)\cdot(1, -i, -j) &= a - bi - cj
\end{aligned}Multiplicações:
\begin{aligned}
z^2 &= (a, b, c)(a, b, c)\\
&= (a + bi + cj)(a + bi + cj)\\
&= a^2 + abi + acj\\
&+ abi - b^2 + bci\\
&+ acj + bcj - c^2\\
&= (a^2 - b^2 - c^2, b(2a + c), c(2a + b))\\
zw &= (a, b, c)(d, e, f)\\
&= (a + bi + cj)(d + ei + fj)\\
&= ad + aei + afj\\
&+ bdi - be + bfi\\
&+ cdj + cej - cf\\
&= (ad - be - cf, ae + b(d + f), af + c(d + e))\\
z \bar{z} &= (a, b, c)(a, - b, - c)\\
&= (a + bi + cj)(a - bi - cj)\\
&= a^2 - abi - acj\\
&+ abi + b^2 - bci\\
&+ acj - bcj + c^2\\
&= (a^2 + b^2 + c^2, - bc, - bc)\\
&=: |z|^2
\end{aligned}Módulo:
\begin{aligned}
|a, b, c|^2 &\leftarrow a^2 + m\circ(b, c) = z \bar{z}\\
&\Rightarrow m\circ(b, c) = b^2 + c^2 - bc(i + j)\\
&\Rightarrow |z|^2 = |a, b, c|^2 = r^2_z - bc(i + j)\\
&\Rightarrow r^2_z = |z|^2 + bc(i + j)
\end{aligned}\begin{aligned}
(i + j)^2 &= (i + j)(i + j)\\
&= i^2 + ij + ji + j^2\\
&= -1 + i + j - 1\\
&= (-2, 1, 1)\\
|(i + j)|^2 &= 2 - (i + j)\\
&= (2, -1, -1)
\end{aligned}\begin{aligned}
[z\bar{z}][w\bar{w}] &= \left[|z|^2\right]\left[|w|^2\right]\\
&= \left[r_z^2 - bc(i + j)\right]\left[r_w^2 - ef(i + j)\right]\\
&= r^2_z r^2_w + bcef(i + j)^2\\
&- r^2_z ef(i + j) - r^2_w bc(i + j)\\
&= r^2_z r^2_w + bcef(-2, 1, 1)\\
&- (r^2_z ef + r^2_w bc)(i + j)\\
&= r^2_z r^2_w - 2bcef\\
&- (r^2_z ef + r^2_w bc - bcef)(i + j)\\
\overline{zw} &= \overline{(ad - be - cf, ae + b(d + f), af + c(d + e))}\\
&= (ad - be - cf, - ae - b(d + f), - af - c(d + e))\\
&= (ad - be - cf, - ae - bd, - af - cd)\\
&- (0, bf, ce)\\
\bar{z}\bar{w} &= (a, -b, -c)(d, -e, -f)\\
&= ad - aei - afj\\
&- bdi - be + bfi\\
&- cdj + cej - cf\\
&= (ad - be - cf, b(f - d) - ae, c(e - d) - af)\\
&= (ad - be - cf, - ae - bd, - af - cd)\\
&+ (0, bf, ce)\\
&= \overline{zw} + 2(0, bf, ce)\\
[zw]\overline{[zw]} &= |zw|^2\\
&= |(ad - be - cf, ae + b(d + f), af + c(d + e))|^2\\
&= (ad - be - cf)^2 + (ae + b(d + f))^2 + (af + c(d + e))^2\\
&- (ae + b(d + f))(af + c(d + e)(i + j)\\
&= r_{zw}^2 - [zw]_2 [zw]_3 (i + j)\\
&= (ad - be - cf, ae + b(d + f), af + c(d + e))\\
&\times (ad - be - cf, - ae - b(d + f), - af - c(d + e))\\
[zw][\bar{z}\bar{w}]\\
\end{aligned}Quando b^2 + c^2 = bc? Suponha que a igualdade vale, e então:
\begin{aligned}
bc &= b^2 + c^2\\
0 &= b^2 - bc + c^2\\
b &= \dfrac{c\pm\sqrt{c^2 - 4c^2}}{2}\\
&= \dfrac{c\pm|c|\sqrt{1 - 4}}{2}\\
&= |c|\dfrac{\hat{c}\pm\sqrt{-3}}{2}\in\R\\
&\iff |c| = 0\\
&\iff b = c = 0
\end{aligned}Como anular uma parte complexa, multiplicando z por um complexo o = (q, w, r) (sua contraparte multiplicativa)? Caso b = c, temos:
\begin{aligned}
zo &=(a + bi + bj)(q + wi + rj)\\
&= aq + awi + arj\\
&+ bqi - bw + brj\\
&+ bqj + bwi - br\\
&= aq - b(w + r)\\
&+ (w(a + b) + bq)i\\
&+ (r(a + b) + bq)j
\end{aligned}Queremos, para definir o, que w(a + b) + bq = r(a + b) + bq = 0:
w(a + b) = r(a + b) = - bq\\
\begin{cases}
\text{Se } a + b = 0: & q = 0\\
\text{Senão}: & w = r = - q\dfrac{b}{a + b}, \text{ com } q \text{ arbitrário.}^1
\end{cases}\\
\therefore o\circ(a, b, b) = \begin{cases}
a + b = 0: &(0, r, w), \text{ com } r, w \text{ arbitrários.}^1\\
a + b\neq 0: &q\left(1 - \dfrac{b}{a + b}(i + j)\right)\\
&= q\left(1 + \left(\dfrac{a}{a + b} - 1\right)(i + j)\right)\\
&= q\dfrac{a + b(- 1 + i + j)}{a + b}
\end{cases}^1 Convenciona-se usar o número 1 para valores arbitrários.
Seja z^\times\leftarrow o\circ |z|^2 tal qual anule multiplicativamente a parte complexa do dividendo em z^-:
\begin{aligned}
z^- &= \dfrac{1}{z}
= \dfrac{1}{z}\cdot\dfrac{\bar{z}}{\bar{z}}
= \dfrac{\bar{z}}{z \bar{z}}
= \dfrac{\bar{z}}{|z|^2}\\
&= \dfrac{a - bi - cj}{a^2 + b^2 + c^2 - bc(i + j)}
\cdot\dfrac{z^\times}{z^\times}
\end{aligned}Se a^2 + b^2 + c^2 - bc = 0:
\begin{aligned}
0 &= a^2 + b^2 + c^2 - bc\\
bc &= a^2 + b^2 + c^2\\
&\iff a = b = c = 0\\
&\iff z = 0\Rightarrow z^-\notin\Complex_3
\end{aligned}Senão, então z^\times = \dfrac{a^2 + b^2 + c^2 - bc(- 1 + i + j)}{a^2 + b^2 + c^2 - bc} e
\text{Seja } r^2 = a^2 + b^2 + c^2, p = bc:\\
\begin{aligned}
z^- &= \dfrac{a - bi - cj}{r^2 - bc(i + j)}
\cdot\dfrac{z^\times}{z^\times}\\
&= \dfrac{
(a - bi - cj)\dfrac{r^2 - bc(- 1 + i + j)}{r^2 - bc}
}{
r^2 + \dfrac{2b^2 c^2}{r^2 - bc}
}\\
&= \dfrac{
\left[(a - bi - cj)\dfrac{r^2 - bc(- 1 + i + j)}{r^2 - bc}\right]
}{
\left[\dfrac{2b^2 c^2 + r^2 (r^2 - bc)}{r^2 - bc}\right]
}\\
&= \dfrac{
(a - bi - cj)(r^2 - bc(- 1 + i + j))
}{
2b^2 c^2 + r^2 (r^2 - bc)
}\\
&= \dfrac{
(a - bi - cj)(r^2 + bc - bci - bcj)
}{
2b^2 c^2 + r^4 - r^2 bc
}\\
&= \dfrac{
(a - bi - cj)(r^2 + p - pi - pj)
}{
2p^2 + r^4 - r^2 p
}\\
&= \dfrac{
ar^2 + ap - api - apj - br^2 i - bp - cr^2 j - cp
}{
2p^2 + r^4 - r^2 p
}\\
&= \dfrac{
a(r^2 + p - pi - pj) - b(p + r^2 i) - c(p + r^2 j)
}{
2p^2 + r^4 - r^2 p
}\\
&= \dfrac{
r^2 (a - bi - cj) + p(a(1 - i - j) - b - c)
}{
2p^2 + r^4 - r^2 p
}\\
&= \dfrac{
(ar^2 + ap - bp - cp, - ap - br^2, - ap - cr^2)
}{
2p^2 + r^4 - r^2 p
}\\
\end{aligned}\begin{aligned}
\vec{z} &\leftarrow (0, b, c) = bi + cj\\
[\vec{z}]^2 &= (bi + cj)(bi + cj)\\
&= - b^2 + bci + bcj - c^2\\
&= - (b^2 + c^2) + bc(i + j)\\
&= - m\circ(b, c)\\
&= - |\vec{z}|^2\\
\vec{z}\vec{w} &= (bi + cj)(ei + fj)\\
&= - be + bfi + cej - cf\\
&= - (be + cf) + bfi + cej\\
\overline{[\vec{z}]}
&= \vec{[\bar{z}]} = -\vec{z}\\
\vec{z}\,\overline{[\vec{z}]}
&= \vec{z}(-\vec{z})
= - [\vec{z}]^2 = |\vec{z}|^2\\
[\vec{z}]^- &= \dfrac{1}{\vec{z}}
= \dfrac{1}{\vec{z}}\cdot
\dfrac{\overline{[\vec{z}]}}{\overline{[\vec{z}]}}
= \dfrac{1}{\vec{z}}\cdot
\dfrac{-\vec{z}}{-\vec{z}}\\
&= \dfrac{-\vec{z}}{-[\vec{z}]^2}
= \dfrac{-\vec{z}}{|\vec{z}|^2}
= -\dfrac{\vec{z}}{m\circ(b, c)}
\end{aligned}Seja \vec{z}^\times tal qual anule multiplicativamente a parte complexa do dividendo em [\vec{z}]^-, ou seja, m\circ(b, c):
\begin{aligned}
\vec{z}^\times &= o\circ m\circ(b, c)\\
&= o\circ(b^2 + c^2, -bc, -bc)\\
&= \begin{cases}
b = c: & (0, 1, 1)\\
b \neq c: & \dfrac{(b^2 + c^2 + bc, - bc, - bc)}{b^2 + c^2 - bc}
\end{cases}
\end{aligned}\begin{aligned}
\hat{z} &\leftarrow \dfrac{\vec{z}}{|\vec{z}|}
= \dfrac{bi + cj}{m^{1\over 2}\circ(b, c)}\\
\hat{z}^2 &= \dfrac{\vec{z}^2}{|\vec{z}|^2}
= \dfrac{-|\vec{z}|^2}{|\vec{z}|^2} = -1\\
\hat{z}\hat{w} &= \dfrac{bi + cj}{m^{1\over 2}\circ(b, c)}
\cdot\dfrac{ei + fj}{m^{1\over 2}\circ(e, f)}\\
&=\dfrac{- be - cf + bfi + cej}{m^{1\over 2}\circ(b, c)\cdot m^{1\over 2}\circ(e, f)}\\
\overline{[\hat{z}]} &= \hat{\bar{z}} = - \hat{z}\\
\hat{z}\bar{\hat{z}} &= \hat{z}(-\hat{z}) = - \hat{z}^2 = -[-1] = 1\rightarrow |\hat{z}|
\end{aligned}m^{1\over 2}\circ(b, c)
Seja q = (l, m, n) : m\circ(b, c) = q^2:
\begin{aligned}
m\circ(b, c) &= (b^2 + c^2, - bc, - bc)\\
&= q^2 = (l, m, n)^2\\
&= (l^2 - m^2 - n^2, m(2l + n), n(2l + m))\\
b^2 + c^2 &= l^2 - m^2 - n^2\\
&\Rightarrow l^2 = b^2 + c^2 + m^2 + n^2\\
- bc &= m(2l + n) = n(2l + m)\Rightarrow\\
-\dfrac{bc}{mn} &= \dfrac{2l}{n} + 1 = \dfrac{2l}{m} + 1\\
&\Rightarrow { m = n\;\circledast^1}\text{ ou }{ l = 0\;\circledast^2 }:
\end{aligned}\circledast^1:
\text{Seja }\vec{r} = r_{\vec{z}}\\
\begin{aligned}
l^2 &= b^2 + c^2 + 2m^2 = \vec{r}^2 + 2m^2\\
\dfrac{-bc}{m^2} &= \dfrac{2l}{m} + 1\\
-bc &= 2lm + m^2\\
0 &= m^2 + 2lm + bc\Rightarrow\\
m &= - l \pm \sqrt{l^2 - bc}\\
m^2 &= l^2 + [l^2 - bc] \mp l\sqrt{l^2 - bc}\\
&= 2l^2 - bc \mp l\sqrt{l^2 - bc}\\
&= l(2l \mp \sqrt{l^2 - bc}) - bc\Rightarrow\\
l^2 &= \vec{r}^2 + 4l^2 - 2bc \mp 2l\sqrt{l^2 - bc}\\
0 &= \vec{r}^2 + 3l^2 - 2bc \mp 2l\sqrt{l^2 - bc}\\
&\Rightarrow [2bc - (\vec{r}^2 + 3l^2)]^2 = 4l^2(l^2 - bc)
\end{aligned}\\
4b^2 c^2 + (\vec{r}^2 + 3l^2)^2 - 4bc(\vec{r}^2 + 3l^2) = 4l^4 - 4l^2 bc\\
4b^2 c^2 + \vec{r}^4 + 9l^4 + 6\vec{r}^2 l^2 - 4bc(\vec{r}^2 + 3l^2) = 4l^4 - 4l^2 bc\\
4b^2 c^2 + \vec{r}^4 + 5l^4 + 6\vec{r}^2 l^2 - 4bc\vec{r}^2 - 12bc l^2 + 4l^2 bc = 0\\
5l^4 + 2l^2(3\vec{r}^2 - 2bc) + 4b^2 c^2 + \vec{r}^4 - 4bc\vec{r}^2 = 0\\
\begin{aligned}
l^2 &= \dfrac{
-(3\vec{r}^2 - 2bc)\pm\sqrt{(3\vec{r}^2 - 2bc)^2 - 5(4b^2 c^2 + \vec{r}^4 - 4bc\vec{r}^2)}
}{5}
\end{aligned}\\
\therefore m^{1\over 2}\circ(b, c) = (l, m, m) = l + [l(2l\mp\sqrt{l^2 - bc}) - bc](i + j)\circledast^2:
\begin{aligned}
l^2 &= b^2 + c^2 + m^2 + n^2 = 0\\
&\Rightarrow b = c = m = n = 0\because b, c, m, n\in\R
\end{aligned}|z| = |a, b, c| = \sqrt{r^2 - bc(i + j)}
Seja q = (l, m, n) : a^2 + m\circ(b, c) = q^2:
\begin{aligned}
|z|^2 &= r^2 - bc(i + j) =\\
q^2 &= (l^2 - m^2 - n^2, m(2l + n), n(2l + m))\\
r^2 &= l^2 - m^2 - n^2\\
l^2 &= r^2 + m^2 + n^2\\
- bc &= m(2l + n) = n(2l + m)\\
- \dfrac{bc}{mn} &= \dfrac{2l}{n} + 1 = \dfrac{2l}{m} + 1\\
&\iff l = 0 \text{ ou } m = n
\end{aligned}Se l = 0:
r^2 = m^2 = n^2 = 0\iff z = 0
Senão, então m = n:
\begin{aligned}
l^2 &= r^2 + 2m^2\\
l &\in \pm\sqrt{r^2 + 2m^2}\\
m &\in \pm\sqrt{l^2 - r^2\over 2}\\
m^2 &= \dfrac{l^2 - r^2}{2}\\
- bc &= m(2l + m) = 2lm + m^2\\
0 &= m^2 + 2lm + bc\\
m &= -l \pm\sqrt{l^2 - bc}\\
m^2 &= l^2 + l^2 - bc \mp l\sqrt{l^2 - bc}\\
&= 2l(l\mp \sqrt{l^2 - bc}) - bc\\
&= \dfrac{l^2 - r^2}{2}
\end{aligned}\\
\begin{aligned}
l^2 - r^2 &= 4l(l\mp \sqrt{l^2 - bc}) - 2bc\\
&= 4l^2\mp 4l\sqrt{l^2 - bc} - 2bc\\
\pm 4l\sqrt{l^2 - bc} &= - l^2 + r^2 + 4l^2 - 2bc\\
&= 3l^2 + r^2 - 2bc\\
16l^2(l^2 - bc) &= (3l^2 + r^2 - 2bc)^2 =\\
16l^4 - 16bc l^2 &= 9l^4 + (r^2 - 2bc)^2 - 6l^2(r^2 - 2bc)\\
&= 9l^4 + (r^2 - 2bc)^2 - 6l^2 r^2 + 12bc l^2\\
0 &= - 7l^4 + 2(14bc - 3r^2)l^2 + (r^2 - 2bc)^2\\
0 &= 7l^4 - 2(14bc - 3r^2)l^2 - (r^2 - 2bc)^2
\end{aligned}\\
\begin{aligned}
l^2 &= \dfrac{(14bc - 3r^2)\pm\sqrt{(14bc - 3r^2)^2 + 7(r^2 - 2bc)^2}}{7}\\
&= 2bc - \dfrac{3r^2\mp\sqrt{196b^2 c^2 + 9r^4 - 84bc r^2 + 7r^4 + 28b^2 c^2 - 28bc r^2}}{7}\\
&= 2bc - \dfrac{3r^2\mp\sqrt{224b^2 c^2 + 16r^4 - 112bc r^2}}{7}\\
&= 2bc - \dfrac{3r^2\mp 4\sqrt{56b^2 c^2 + r^4 - 28bc r^2}}{7}
\end{aligned}e^x
\begin{aligned}
exp\circ z &= e^z\\
&= e^{(a, b, c)}\\
&\leftarrow e^{a + bi + cj}\\
&= e^a e^{bi} e^{cj}\\
&= e^a (\cos b + i \sin b) (\cos c + j \sin c)\\
&= e^a (\cos b \cos c + j \cos b \sin c + i \sin b \cos c + i \sin b \sin c)\\
&= e^a (\cos b \cos c + j \cos b \sin c + i (\sin b (\cos c + \sin c))\\
&= e^a (\cos b \cos c, \sin b (\cos c + \sin c), \cos b \sin c)\\
e^{a + cj + bi}
&= e^a (\cos b \cos c, \sin b \cos c, \sin c (\cos b + \sin b))
\end{aligned}Operações
Que operação k\circ z leva a + bi à a + bj e vice-versa?
Seja z = a + bi + cj:
\begin{aligned}
k\circ z &= k\circ(a + bi + cj) = a + ci + bj\\
a + ci + bj &= \Re(z) + \Im((i + j)\Im(z))\\
&= \Re(z) + \Im((i + j)(bi + cj))\\
&= a + \Im(- b + ci + bj - c)\\
&= a + ci + bj\\
\end{aligned}\\
\therefore k\circ z = \Re(z) + \Im((i + j)\Im(z))