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I was reading up on how to find the square root of i , and I learned that multiplication of complex numbers could be viewed geometrically by viewing the complex numbers as coordinates on the complex number plane $a_1+b_1i = (a_1,b_1)$ and $a_2+b_2i = (a_2,b_2)$. One can take the polar coordinates of the complex numbers to give $(a_1,b_1) \Rightarrow r_1$, angle = $w$ and $(a_2,b_2) \Rightarrow r_2$, angle = $k$ . And finally the multiplication of the two numbers can be viewed as multiplying $r_1$ and $r_2$, while adding the angles $w$ and $k$, to give the product $r_3= (r_1)(r_2)$, angle = $w+k$ . This can be used to intuitively find the square root of $0 + 1i$.

However in my math textbooks I have not seen any type of vector multiplication similar to this in regards to the real numbers, only dot products and cross products. Does this type of multiplication serve some purpose in regards to real number vectors, does it describe something interesting? Or is it only useful when it comes to multiplying vectors in the complex number plane?

Edit: $r$ is describing the overall length, or magnitude, of the vector. The angle represents the direction the vector is pointing in in regards to the plane it's on.

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  • $\begingroup$ Related question: math.stackexchange.com/q/2296374/221811 $\endgroup$ – Chappers Jun 7 '17 at 22:46
  • $\begingroup$ It has to do with dimensions; real numbers are 1-D, complex numbers are 2-D. See De Moivre's Theorem which relates to your comment on rectangular vs. polar coordinates. See also quaternions which are useful although not as directly related to "reality" since complex numbers are sufficient to have a closed field of values under the normal arithmetic operations (multiplication, division, addition, subtraction, exponentiation, etc.) $\endgroup$ – Jason S Jun 8 '17 at 17:40
  • $\begingroup$ Viewing $\mathbb{R}^2$ as $\mathbb{C}$ gives a commutative, associative multiplication map $\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}^2$. One way to make your question precise is to ask whether there are analogous maps (perhaps not so nice, e.g. perhaps not commutative, but still associative) $\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n$ when $n>2$. For $n=4$ the answer is yes, if you give up commutativity: you can identify $\mathbb{R}^4$ with the space of quaternions $\mathbb{H}$, but that multiplication is no longer commutative. For other $n$ the answer is no, surprisingly. $\endgroup$ – symplectomorphic Jun 9 '17 at 2:09
  • $\begingroup$ (This is due to the Frobenius theorem on real division algebras.) Hamilton discovered the quaternions after searching in vain for a similar multiplication when $n=3$. $\endgroup$ – symplectomorphic Jun 9 '17 at 2:09
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When doing vector geometry in the plane (pure vector geometry with no coordinate system), all directions “look equal”.

Introducing complex multiplication of such vectors requires a breaking of this symmetry: you must single out a direction to play the role of the positive real axis, so that you can define the polar angle. And then vectors in different directions will behave differently with respect to multiplication. Think of squaring, for example: $1$ maps to itself ($1^2=1$), while $i$ doesn't ($i^2=-1 \neq i$).

The dot and cross products are more geometrical, in the sense that they don't depend on making certain directions special. (Well, actually the cross product is rather weird too, and it's better to learn about the exterior and Clifford products, as recommended in the answer by Bye_World.)

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  • $\begingroup$ Fixing the data required to specify a dot product or cross product also breaks symmetry. In a vector space $\Bbb V$, for any nonzero vectors $v, v'$ there is a vector space isomorphism $\Bbb V \to \Bbb V$ that maps $v$ to $v'$. But if we introduce an inner product $\cdot$ on $\Bbb V$, the same is not true for isometries (isomorphisms of the inner product space): We can only find an isometry iff $v \cdot v = v' \cdot v'$. For the cross product in $\Bbb R^3$, we require an orientation-preserving isometry. $\endgroup$ – Travis Willse Jun 8 '17 at 0:26
  • $\begingroup$ We can formalize "they don't depend on making certain directions special" in this language, too: The group of isometries of $\cdot$, which is isomorphic to $O(n,\Bbb R)$ acts transitively on the space $\Bbb P(\Bbb R^n)$ of lines in $\Bbb R^n$---so every direction is on equal footing. On the other hand, the only isomorphisms $\Bbb R^2 \to \Bbb R^2$ preserving complex multiplication are the identity and complex conjugation. The induced action of this space on $\Bbb P(\Bbb R^2)$, however, is not transitive; the preferred direction (the real axis) is one of its fixed points of the action. $\endgroup$ – Travis Willse Jun 8 '17 at 0:34
  • $\begingroup$ @Travis: You're completely right, but I avoided this in order to keep things a bit simpler. I guess it also depends on what one means by “vector geometry in the plane” to begin with. In elementary courses, it's usually(?) taken for granted that one has the concepts of length and angle, and then the inner product is defined in terms of these (instead of the other way around): $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\alpha$. In that case it makes sense to say that the inner product respects the symmetry of the plane. $\endgroup$ – Hans Lundmark Jun 8 '17 at 6:39
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The analogy you’re looking for isn’t any type of vector multiplication per se. It’s a matrix multiplication. Identify the complex number $z=a+bi=re^{i\theta}$ with the matrix $$Z=\pmatrix{a&-b\\b&a}=r\pmatrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta},$$ a scaled rotation. Then the product $z_1z_2$ directly maps to the matrix product $Z_1Z_2$. This identification corresponds to the geometric interpretation of multiplication by a complex number as a combination rotation and dilation.

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Let $I=e_1\wedge e_2$ be the positively-oriented unit pseudoscalar of the plane. Then $a+bI$ where $a,b\in\Bbb R$ is a spinor$^\dagger$ -- i.e. an object that we use to rotate (and scale) vectors and their higher dimensional analogs. Then the composition of spinors corresponds to multiplication -- where multiplication works exactly as it does with the complex numbers: $$R_1 = a+bI \\ R_2 = c+dI \\ R_2\circ R_1 = (c+dI)(a+bI) = (ca-db)+(cb+da)I$$

Note that we can also represent spinors in polar form: $$R_1 = a+bI = |r_1|e^{\theta_1I} \\ R_2 = c+dI = |r_2|e^{\theta_2I} \\ R_2\circ R_1 = |r_1||r_2|e^{(\theta_1+\theta_2)I}$$

So algebraically, the set of spinors in the plane behave exactly like complex numbers. And, in fact, moving up a dimension, the set of spinors in $3$-space behave exactly like quaternions.

References

For detailed information on the mathematics behind spinors/ rotors, the Wikipedia page on Geometric Algebra is fairly thorough. However more comprehensive references include

  • Linear and Geometric Algebra by Alan Macdonald
  • Clifford Algebra to Geometric Calculus by David Hestenes & Garret Sobczyk
  • Geometric Algebra for Physicists by Chris Doran & Anthony Lasenby

$\dagger:$ If $a^2+b^2 = 1$ -- that is, if the spinor is normalized -- then the action of it on a vector is simply to rotate rather than rotate and scale and in this special case they are called rotors.

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This special multiplication works for complex numbers, i.e. is good enough only for $\mathbb{R}^2 \equiv \mathbb{C}$, but is non-trivial to generalize to make useful for beginners :)

The dot product is generalizable and the cross product, although is defined only in $\mathbb{R}^3$, but has lots of useful applications in physics, and hence is covered in the beginner math classes...

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The method you describe isn't related to any vector interpretation of complex numbers.

It comes from the fact that the complex number $z = a + bi$ can be written in so-called polar form $re^{i\theta}$, where $r = |z| = \sqrt{a^2+b^2}$ is the modulus of $z$ and $\theta$ is the "argument" of $z$, which is basically the angle that the point $(a,b)$ makes with the positive $x$-axis (e.g. the point $(1,1)$, corresponding to $z = 1+i,$ has $\theta = \pi/4$).

If you have two complex numbers in polar form, say $z = r_1e^{i\theta_1}$ and $w = r_2e^{i\theta_2}$, then $$zw = r_1e^{i\theta_1} \cdot r_2e^{i\theta_2} = r_1 \cdot r_2 \cdot e^{i\theta_1 + i\theta_2} = r_1r_2 e^{i(\theta_1+\theta_2)}$$

This is a complex number whose magnitude (modulus) is $r_1r_2$ and whose argument is $\theta_1 + \theta_2$.

So while the polar form can be viewed as a description of a vector (magnitude and direction), this vector interpretation is not where the $r_1r_2$ and $\theta_1 + \theta_2$ come from. $r_1r_2$ comes from multiplication as usual and $\theta_1 + \theta_2$ comes from multiplying exponential expressions with the same base.

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Put another way, the product $m : \Bbb R^2 \times \Bbb R^2 \to \Bbb R^2$ you describe is simply complex multiplication $\Bbb C \times \Bbb C \to \Bbb C$ written in terms of its real and imaginary parts, that is, via the usual identification $\Bbb C \leftrightarrow \Bbb R^2$. Explicitly, it is $$m((x_1, y_1), (x_2, y_2)) := (x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1) .$$

We can actually write $m$ without too much trouble in terms of the usual dot and cross product, so if one takes those operations to be primitive, $m$ is not really new (though in general I'd consider complex multiplication to be a primitive object). To describe this relationship, we need two ingredients:

(1) Via the above identification, complex conjugation $z \mapsto \bar z$ induces a map $\bar{\cdot} : (x, y) \mapsto (x, -y)$ on $\Bbb R^2$.

(2) The usual cross product determines a map $s : \Bbb R^2 \times \Bbb R^2$: If we regard $\Bbb R^2$ as a subset of $\Bbb R^3$ via the embedding $\Phi: (x, y) \mapsto (x, y, 0)$, computing gives that the cross product $\Phi(x_1, y_1) \times \Phi(x_2, y_2)$ is $(0, 0, x_1 y_2 - x_2 y_1)$, defining a skew-linear map $s((x_1, y_1), (x_2, y_2)) := x_1 y_2 - x_2 y_1$. Put another way, if we regard ${\bf x}_a := (x_a, y_a)$ as column vectors, $s({\bf x}_1, {\bf x}_2) = \det\pmatrix{{\bf x}_1 & {\bf x}_2}$, so this quantity is the (signed) area of the parallelogram with sides ${\bf x}_1$ and ${\bf x}_2$.

With this in hand, we can see that the map $m$ can be written in terms of the dot product on $\Bbb R^2$ and, via $s$, the cross product on $\Bbb R^3$, as $$m({\bf x_1}, {\bf x}_2) = (\overline{{\bf x}_1} \cdot {\bf x}_2, s(\overline{{\bf x}_1}, {\bf x}_2)) .$$ In short, the real and imaginary components of the map $m$ are (up to composition with a conjugation) respectively the dot product and the cross product---$m$ just packages these two maps together in a particular way.

The situation in which one defines an operation on $\Bbb R^4$ via the identification of the quaternions, $\Bbb H$, with $\Bbb R^4$ can be handled similarly.

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I have a copy of a paper that explores alternate product structures on ordered pairs of Reals. The basic rules are: (a) Addition is vector addition, i.e (a,b) + (c,d) = (a+c,b+d); (b) Multiplication by 'real-part' pairs acts like scalar multiplication, i.e (a,0) * (b,c) = (b,c) * (a,0) = (ab,ac); (c) Multiplication distributes over addition in the usual manner. To specify a particular structure, one must give constants p, q, and define (0,1) * (0,1) = (p,q), which yields the full product rule (a,b) * (c,d) = (ac+pbd,bc+ad+qbd). Note that the complex numbers correspond to (p,q) = (-1,0). This construction yields an integral domain for any choice of p and q, and a field for some choices [sorry,forgot the exact criterion - I do recall it is 'discriminant'-related...]

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    $\begingroup$ "I have a copy of a paper that..." Please cite the author and publication details for this paper. $\endgroup$ – hardmath Jun 7 '17 at 23:57
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The problem with this type of multiplication for general vectors is that it only works in 2 dimensions. In 1D, there is only one direction, so it is the same as scalar multiplication. In Higher than 2D, there are many axes you could rotate a vector around.

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