Technologically Embodied
Geometric Functions
Vector Multiplication on the Complex Plane

Return to the complex plane unit description.

In this activity you will use the complex plane to explore an interesting and revealing approach to vector multiplication. The approach here is inspired by Tristan Needham [See Needham's Visual Complex Analysis, p. 27].

The relationship between vector operations and the complex plane is an obvious one in several respects. Complex numbers can be considered either as points in the complex plane or as vectors. When complex numbers are considered as vectors, complex addition is identical to vector addition.

This relationship extends to vector multiplication as well. Not only is addition in the complex plane the same as vector addition, but vector multiplication can be expressed in the complex plane in a striking and useful way. In this visualization the vector dot product (also known as the scalar product) will appear as a horizontal (real) quantity

In physics the vector cross product is represented in a third direction, perpendicular to the plane of the two vectors being multiplied. But in only two dimensions we must represent the cross product differently, so we redefine the direction of `a ⨯ b` as being in the direction of either the positive or the negative imaginary axis depending on the sine of the angle from `a` to `b`.

In this representation, then

  • The dot product is a real value: `a · b = r_a r_b cos(theta_b - theta_a)`.
  • The cross product is an imaginary value: `a ⨯ b = i r_a r_b sin(theta_b - theta_a)`.

In the blank websketch below, construct two vectors `a` and `b`, and measure their polar coordinates. Use the Triangle 1 tool to represent the projection of `b` on `a`.

The projection is of length `r_b cos (theta_b - theta_a)`. To show the dot product `a · b` in the horizontal (real) direction, we must rotate the projection to the real axis (rotating by `-theta_a`) and dilate it by `r_a`.

We can accomplish both at once by multiplying 'b' by the conjugate of `a`. (Why does this work?)

Use Triangle 2 to show the resulting image of the shaded triangle. Why is the base of this triangle the real number `a · b`? Why is its height the imaginary number `a ⨯ b`?


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Update History:

18 October 2016: Created this page.