38 Chapter^1
we proceed as follows (see Fig. 1.7). Suppose that OA is the vector corre-
sponding to the number (1, 0), and consider the triangle OAP. Choosing
OQ as the side corresponding to OA, we construct a triangle OQM similar
to OAP with the side OM making an angle 01 + 02 with OX. Then the
vector OM corresponds to the product z 1 z 2 • In fact, by the construction,
OM has the appropriate argument, i.e., 01 + 02 , and as to its modulus
we have, by the similarity of triangles OQM and OAP, r/r1 = r2/l, so
that r = r1r2.
( d) Division. To find the position vector OD of the quotient,
z = z2 = r2 ei(02-0i)
Z1 r1
where z 1 ::j=. 0, z 2 ::j=. 0, a construction analogous to the preceding one can
be made, except that the triangle OQD similar to OAP is constructed in
such a manner as to have the side OD forming an angle 02 - 01 with OX
(Fig. 1.8). Then, by part (c), we have z1z = z2, or z = z2/z1.
Remark If 01 = 0 or 01 = 7r, the triangle OAP collapses. Then z 1 is
real and the vector OD has the same orientation as OQ if 01 = 0, and
the opposit~ orientation if 01 = 7r. As to the length r of OD, it satisfies
r/1 = r 2 /r 1 , so that it can be constructed as a fourth proportional between
OA, OQ, and OP by the usual method of elementary geometry (Fig. 1.9).
The same remark applies to the construction in part (c). However, if one
of the factors is nonreal, we can use the corresponding position vector to
form a triangle with OA and then proceed with the construction as before.
Finally, we wish to consider some special constructions for z = 1 / z1
(z 1 ::j=. 0), i.e., for the reciprocal of a complex number. Clearly, lzllz 11 =
rr 1 = 1, and Arg z = - Arg z 1. Hence z is the inverse of z 1 with respect
to the circle of radius 1 with center at the origin (the so-called unit circle).
y
Qx
Fig. 1.8