Showing that the distributive law holds in twizzle arithmetic is equivalent to proving the trigonometric addition formulae.

\begin{matrix} r_{1} cis (θ_{1}) \times r_{2} cis (θ_{2}) & = & r_{1} r_{2} cis (θ_{1} + θ_{2}) \\ = & r_{1} r_{2} (\cos θ_{1} + i \sin θ_{1}) (\cos θ_{2} + i \sin θ_{2}) \\ = & r_{1} r_{2} ((\cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2}) + i (\sin θ_{1} \cos θ_{2} + \cos θ_{1} \cos θ_{2})) \end{matrix}

That's not the focus of this question though. What we're really looking for is some experimentation with the animation leading to an understanding of the behaviour of (z-i) , (z+i) , and (z-i)(z+i) as z moves in a loop-like path.

It's quite useful to look at the case where the beige a -twizzle is zero in the second animation.