Exercise [05.02]

dickdock · 25 Jul 2008, 01:38

Exercise [05.02] Check the parallelogram law (complex addition) and the similar-triangle law (complex multiplication) by trigonometry and direct computation.

In general, if A, B are any points with coordinates (x_a, y_a), (x_b, y_b)

, then the length of $\overline{AB} = \sqrt{(x_b - x_a)^2 + (y_b - y_a)^2}$

Parallelogram law

Let P = W + Z

. OWPZ is a parallelogram if opposite sides have the same lengths.

O = (0, 0), W = (x_w, y_w), P = (x_w + x_z, y_w + y_z), Z = (x_z, y_z)

Length of $\overline{OW} = \sqrt{(x_w - 0)^2 + (y_w - 0)^2} = \sqrt{(x_w + x_z - x_z)^2 + (y_w + y_z - y_z)^2}$ = length of $\overline{ZP}$
Length of $\overline{WP} = \sqrt{((x_w + x_z) - x_w)^2 + ((y_w + y_z) - y_w)^2} = \sqrt{(x_z - 0)^2 + (y_z - 0)^2}$ = length of $\overline{OZ}$

Hence OWPZ is a parallelogram.

Similar-triangle law

Let P = WZ

. Then

P = (x_p, y_p) = (x_wx_z - y_wy_z, x_wy_z + y_wx_z)

Triangle OZP is similar to triangle O1W if the ratios of the lengths of corresponding sides are the same, i.e. if $\frac{\overline{OZ}}{\overline{O1}} = \frac{\overline{ZP}}{\overline{1W}} = \frac{\overline{PO}}{\overline{WO}}$

The length formula throws up some tedious expressions. To ease simplification of these we note first that |wz|^2 = (|w||z|)^2

:

x_p^2 + y_p^2 = (x_wx_z - y_wy_z)^2 + (x_wy_z + y_wx_z)^2

= x_w^2x_z^2 - 2x_wx_zy_wy_z + y_w^2y_z^2 + x_w^2y_z^2 + 2x_wx_zy_wy_z + y_w^2x_z^2

= x_w^2x_z^2 + y_w^2y_z^2 + x_w^2y_z^2 + y_w^2x_z^2

and that

(x_p - x_z)^2 + (y_p - y_z)^2 = x_p^2 - 2x_px_z + x_z^2 + y_p^2 - 2y_py_z + y_z^2

= x_p^2 + y_p^2 - 2(x_wx_z - y_wy_z)x_z + x_z^2 - 2(x_wy_z + y_wx_z)y_z + y_z^2

= x_p^2 + y_p^2 - 2x_wx_z^2 + x_z^2 - 2x_wy_z^2 + y_z^2

= x_p^2 + y_p^2 + x_z^2(1 - 2x_w) + y_z^2(1 - 2x_w)

= (x_w^2 + y_w^2)(x_z^2 + y_z^2) + (x_z^2 + y_z^2)(1 - 2x_w)

= (x_w^2 - 2x_w + 1 + y_w^2)(x_z^2 + y_z^2)

The ratios are

$\frac{\overline{OZ}}{\overline{O1}} = \frac{\sqrt{(x_z - 0)^2 + (y_z - 0)^2}}{\sqrt{(1 - 0)^2 + (0 - 0)^2}} = \sqrt{x_z^2 + y_z^2}$

$\frac{\overline{PO}}{\overline{WO}} = \frac{\sqrt{(0 - x_p)^2 + (0 - y_p)^2}}{\sqrt{(0 - x_w)^2 + (0 - y_w)^2}} = \frac{\sqrt{ (x_w^2 + y_w^2)(x_z^2 + y_z^2)}}{\sqrt{x_w^2 + y_w^2}} = \sqrt{x_z^2 + y_z^2}$

$\frac{\overline{ZP}}{\overline{1W}} = \frac{\sqrt{(x_p - x_z)^2 + (y_p - y_z)^2}}{\sqrt{(x_w - 1)^2 + (y_w - 0)^2}} = \frac{\sqrt{(x_w^2 - 2x_w + 1 + y_w^2)(x_z^2 + y_z^2)}}{\sqrt{x_w^2 - 2x_w + 1 + y_w^2}}$
$= \sqrt{x_z^2 + y_z^2}$

Hence the triangles are similar, the scaling factor being |Z|. There has to be a better way.

Orientation

Complex multiplication also preserves orientation. One way to define orientation (taken from the excellent 'Geometry Unbound' book by Kiran Kedlaya, at http://www-math.mit.edu/~kedlaya/geometryunbound/) is via signed areas:

If P_1, P_2, P_3

are the vertices of a triangle, then the signed area is defined as
$[P_1P_2P_3]_{\pm} = \frac{1}{2}(x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3)$

This is known as the surveyor's or shoelace formula (and applies to polygons of any number of vertices), since if the coordinates are laid out in a 2x(n + 1) array (plus 1 for the repeat of the first coords at the end) with the x's at the top and the y's at the bottom, the products are crisscrossed along much like one does with a shoelace, or a surveyor.

Orientation is then defined to be the sign of the signed area, positive being counter-clockwise, negative clockwise.

For triangle O1W with coords (0, 0), (1, 0), (x_w, y_w)

$[O1W]_{\pm} = \frac{1}{2}(0 - 0 + 1.y_w - 0 + 0 - 0) = \frac{1}{2}(y_w)$

For triangle OZP with coords (0, 0), (x_z, y_z), (x_zx_w - y_zy_w, x_zy_w + y_zx_w)

$[OZP]_{\pm} = \frac{1}{2}(0 - 0 + x_z(x_zy_w + y_zx_w) - (x_zx_w - y_zy_w)y_z + 0 - 0)$
$= \frac{1}{2}(x_z^2y_w + x_zy_zx_w - x_zx_wy_z + y_z^2y_w) = \frac{1}{2}(x_z^2 + y_z^2)y_w$

Since

is always positive, the orientation of OZP depends only on y_w

, the same as O1W. Hence complex multiplication preserves orientation.

Exercise [05.02]

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