1550251515-Classical_Complex_Analysis__Gonzalez_

36 Chapter^1

the segment OC by OC, that of OB by OB, and so on, we have

OC = OB+ BC= OB+ OA =a+ c

CR= CD+DR= BP+AQ = b+d

which proves that the vector OR represents the complex number (a +

c, b + d).

Example The sum of the complex numbers ( 4, 1) and ( -2, 2) is shown

geometrically in Fig. 1.5.

We note that the inequality

follows at once geometrically from Fig. 1.4, the < sign corresponding to

the diagram on the left (triangle inequality) and the= sign corresponding

to the diagram on the right (collinear vectors with the same orientation).

The < sign also corresponds to the case where the vectors are collinear but

with opposite orientations (not shown in the figure).

(b) Subtraction. We have seen that

(a, b) - ( c, d) = (a, b) + ( -c, -d)

Hence, to subtract geometrically the number ( c, d) from (a, b ), it suffices

to add to the vector OP representing (a,b) the vector OQ' representing

(-c, -d), i.e., the symmetric vector of OQ with respect to the origin (as

shown in Fig. 1.6).

It should be noted that if the vectors OP and OQ are not collinear, the

vector OS representing the difference (a - c, b - d) is equipolent (equal

in length and parallel) to the vector QP, which coincides in magnitude

and direction with the second diagonal of the parallelogram OP RQ, this

diagonal being considered as a directed segment with origin at Q.

x

Fig. 1.fi