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Given $$e^{\textbf{i}\theta}=cos(\theta) + \textbf{i}sin(\theta) \ \ \ \ (1)$$ and $$e^{-\textbf{i}\theta}=cos(\theta) - \textbf{i}sin(\theta) \ \ \ \ (2)$$ To find $sin(\theta)$ first you should perform:

$$\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=\textbf{i}sin(\theta) \ \ \ \ (3)$$
The seemingly obvious thing to do would then be to just "divide" the $\bf{i}$ over to get $\frac{1}{\textbf{i}}\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2\textbf{i}}$ (which is the answer I'm working toward). But that introduces a scalar (1) dividing a bivector ($\textbf{i}$). How can you just divide a scalar by a bivector?

UPDATE

Despite the help given here in the comments, I've managed to find an answer. It is a simple answer:

Following $(3)$, the next step would be to product both sides of the equation by $-\textbf{i}$ on the left giving $$-\textbf{i}\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=sin(\theta) \ \ \ \ (4)$$ Then all there is to note is that $\frac{1}{\textbf{i}}$ is simply an abbreviation for $-\textbf{i}(=\textbf{i}^{-1})$. With that in mind, $(4)$ becomes $$\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2\textbf{i}}=sin(\theta) \ \ \ \ (5)$$

roshoka
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  • The set $\mathbb C\setminus{0}$ is a group for the $\times$ operation and $i^{-1}$ is just the inverse of $i$ for $\times$. The neutral element for $\times$ is $1$ hence it suffices to check that $i\times(-i)=1$... and surely you can do that? – Did Jan 26 '19 at 18:57
  • You can definitely do the $\textbf{i(-i)}=1$, I'm referring more to the $\frac{1}{\textbf{i}}$ part – roshoka Jan 26 '19 at 19:13
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    ?? We do not "define $i^{-1}$ as $-i$, we check that $i^{-1}=-i$. – Did Jan 26 '19 at 19:14
  • I'll edit it to make it more clear. – roshoka Jan 26 '19 at 20:02
  • Your question makes no sense. Complex numbers are far more than a vector space over the reals. Please look up any actual definition of complex numbers before you pose silly challenges like "how can you just ...". – user21820 Feb 06 '19 at 16:48
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    I'm not talking about $\mathbb{C}$ here. There is no complex plane in geometric algebra. A GA complex number is defined as any member of $\mathbb{G}^n$ that can be represented as $a+b\textbf{i}$ where $\textbf{i}$ is a bivector resulting from the geometric product of two vectors. The question was simple: What does it mean to divide a scalar 1 by a bivector $\textbf{i}$? I've since learned that it is simply a notational thing. $1/\textbf{i}$ is simply an abbreviation for $-\textbf{i}$ (again, which is a bivector, not a part of the traditional complex plane). Thanks for all of your help. – roshoka Feb 06 '19 at 23:31
  • Even if you cannot divide, then multiply both sides by $i$. You should edit your question to make it clear that your $i$ is not the complex square-root of $-1$. – user21820 Feb 07 '19 at 14:54
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    I think the "Geometric Algebra" in the title and the "geometric-algebras" tag should be sufficient to show that this isn't dealing with $\mathbb{C}$. Either way, I have the answer now. I'm going to make an edit to the question showing the answer, since I can't answer it myself due to the closure. – roshoka Feb 07 '19 at 15:29
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    I think, as a general rule going forward, people need to look up any actual definition of complex numbers in geometric algebra before posing responses. – roshoka Feb 07 '19 at 15:43

1 Answers1

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$\mathbf{B}^{-1}$ for bivector $\mathbf{B}$, is just a bivector, and differs from $\mathbf{B}$ by a scalar factor. When $\mathbf{i}$ is a unit bivector satisfying $\mathbf{i}^2 = -1$, you can see by inspection that $\mathbf{i}^{-1} = -\mathbf{i}$.

To gain a bit more insight into why this works, consider the fact that you can always factor a bivector $ \mathbf{i} $ into a product of orthogonal vectors $ \mathbf{u} \mathbf{v} $, (at least in 2-3 dimensions), after which you can invert the bivector as follows $$ \mathbf{i}^{-1} = \frac{1}{\mathbf{v}} \frac{1}{\mathbf{u}} = \frac{\mathbf{v} \mathbf{u}}{\mathbf{u}^2 \mathbf{v}^2}.$$

The simplest examples are those where the bivector is some scalar multiple of two clearly orthonormal factors, such as

$$(\mathbf{e}_1 \mathbf{e}_2)^{-1} = \frac{1}{{\mathbf{e}_2}} \frac{1}{{\mathbf{e}_1}} = \mathbf{e}_2 \mathbf{e}_1,$$

A less trivial example is $ \mathbf{i} = \frac{1}{{\sqrt{3}}}\left( { \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_2 \mathbf{e}_3 + \mathbf{e}_3 \mathbf{e}_1 } \right) $, where possible factorizations include $$\begin{aligned} \mathbf{i} &= \frac{1}{{\sqrt{3}}} \left( { \mathbf{e}_1 + \mathbf{e}_2 - 2 \mathbf{e}_3 } \right) \frac{ \mathbf{e}_2 - \mathbf{e}_1 }{2} \\ &= \frac{1}{{\sqrt{3}}} \frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2} \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right),\end{aligned}$$ but once you find such factors, you can easily compute $\mathbf{i}^{-1} = -\mathbf{i}$ using these factors, for example:

$$\begin{aligned}\mathbf{i}^{-1} &= 2 \sqrt{3} \frac{ 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 }{6} \frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2 } \\ &= \frac{1}{{2 \sqrt{3}}} \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right) \left( { \mathbf{e}_3 - \mathbf{e}_2 } \right) \\ &= \frac{1}{{\sqrt{3}}} \left( { \mathbf{e}_1 \mathbf{e}_3 + \mathbf{e}_2 \mathbf{e}_1 + \mathbf{e}_3 \mathbf{e}_2 } \right) \\ &= -\mathbf{i}.\end{aligned}$$

Peeter Joot
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