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Complex numbers via rigid motions Just a bit mathematical ! Rotations around the origin. 180 deg or pi Y = -y X = -x so (1,0) goes to (-1,0) Let us call this transformation H (for a half turn) 90 deg or pi/2 Y = x X = -y so (1,0) goes to (0,1) and (-1,0) goes to (0,-1) and (0,1) goes to (-1,0) and (0,-1) goes to (1,0) Let us call this transformation Q (for a quarter turn) Then H(x,y) = (-x,-y) and Q(x,y) = (-y,x) We know that H(H(x,y)) = (x,y) and if we are bold we can write HH(x,y) = (x,y) and then HH = I, where I is the identity or do nothing transformation. In the same way we reach QQ = H Now I is like multiplying the coodinates by 1 and H is like multiplying the coordinates by -1 This is not too outrageous as a dilation can be seen as a multiplication of the coordinates by a number <> 1 So, continuing into uncharted territory, we have H squared = 1 (fits with (-1)*(-1) = 1 and Q squared = -1 (fits with QQ = H, at least) So what is Q ? Let us suppose that it is some sort of a number, and let its value be called k. What we can be fairly sure of is that k does not multiply each of the coordinates. This is most likely only for the normal numbers. Now the "number" k describes a rotation of 90, so we would expect that the square root of k to describe a rotation of 45 At this point it helps if the reader is familiar with extending the rational numbers by the introduction of the square root of 2 (a surd, although this jargon seems to have disappeared). Let us assume that sqrt(k) is a simple combination of a normal number and a multiple of k: sqrt(k) = a + bk Then k = sqr(a) + sqr(b)*sqr(k) + 2abk, and sqr(k) = -1 which gives k = sqr(a)-sqr(b) + 2abk and then (2ab-1)k = sqr(a) - sqr(b) From this, since k is not a normal number, 2ab = 1 and sqr(a) = sqr(b) which gives a = b and then a = b = 1/root(2) Now we have a "number" representing a 45 degree rotation. namely (1/root(2)*(1 + k) If we plot this and the other rotation numbers as points on a coordinate axis grid with ordinary numbers horizontally and k numbers vertically we see that all the points are on the unit circle, at postions corresponding to the rotation angles they describe. OMG there must be something in this ! ! !