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Section 1.4 The Geometry of Complex Numbers, Continued

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COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/

CHAPTER 1 COMPLEX NUMBERS

Section 1.4 The Geometry of Complex Numbers, Continued

    In Section 1.3 we saw that a complex number could be viewed as a vector in the xy-plane whose tail is at the origin and whose head is at the point (x,y). A vector can be uniquely specified by giving its magnitude (i.e., its length) and direction (i.e., the angle it makes with the positive x-axis). In this section, we focus on these two geometric aspects of complex numbers.

    Let be the modulus of (i.e., ), and let be the angle that the line from the origin to the complex number makes with the positive x -axis. (Note: The number is undefined if . Then

(1-25) .

Definition 1.9: Polar Representation

The identity = ( ) = is known as a polar representation of , and the values and are called polar coordinates of .

Example 1.7, Page 23. Find several polar forms of .

z1 := 1 + I:
z2 := sqrt(2)*cos(Pi/4) + I*sqrt(2)*sin(Pi/4):
z3 := sqrt(2)*cos(9*Pi/4) + I*sqrt(2)*sin(9*Pi/4):
z4 := sqrt(2)*cos(-7*Pi/4) + I*sqrt(2)*sin(-7*Pi/4):
`z ` = z1; ` `;
`A few polar forms for z.`;
`sqrt(2)cos(Pi/4) + i sqrt(2)sin(Pi/4)` = z2;
`sqrt(2)cos(9Pi/4) + i sqrt(2)sin(9Pi/4)` = z3;
`sqrt(2)cos(-7Pi/4) + i sqrt(2)sin(-7Pi/4)` = z4;

Definition 1.10:

If , then .

If , we say that is an argument of .

An argument of is or provided that .
The exponential form of is , where and .

Example 1.8, Page 24. Because , we have

Definition 1.11:

Let be a complex number. Then

, provided and < .

If = , we say that is the argument of .

Example 1.9, Page 24. .

Example 1.10, Page 24. Find the polar form of , by computing and .

>	z1 := - sqrt(3) - I: `z1 ` = z1; ` `; r := abs(z1): t := argument(z1): ` r ` = r, theta = t; z2 := r(cos(t) + Isin(t)): `z2 = 2cos(-5Pi/6) + i 2sin(-5Pi/6)` = z2; ` `; `Does z1 = z2 ?`; z1 = z2; evalb(z1 = z2);

Example 1.11, Page 25. Write in the form.

>	z1 := 4I: `z1 ` = z1; ` `; r := abs(z1): t := argument(z1): ` r ` = r, theta = t; z2 := rexp(I*t): `z2 = 4 exp(iPi/2)` = z2; ` `; `Does z1 = z2 ?`; z1 = z2; evalb(z1 = z2);

Example 1.12, Page 26. Given , find and .

>	Z := 1 + I: `z ` = Z; z:='z': R := abs(Z): `r ` = R; r:='r': ` `; cz := conjugate(Z): w1 := 1/R^2*conjugate(Z): w2 := 1/Z: conjugate(z) = cz; ` `; conjugate(z)/r^2 = w2; `1/z ` = w2;

Theorem 1.3 If = and = , then as sets

Example 1.13, Page 28. Given and , compute using polar computations.

z1 := 8*I: `z1 ` = z1;
r1 := abs(z1):
t1 := argument(z1):
z1 = r1*exp(I*t1):
`z1 = 8 exp(iPi/2)` = z1; ` `;
z2 := 1 + I*sqrt(3): `z2 ` = z2;
r2 := abs(z2):
t2 := argument(z2):
z2 = r2*exp(I*t2):
`z2 = 2 exp(iPi/3)` = z2; ` `;
w1 := z1/z2:
`w1 = z1/z2 ` = w1;
w1 := evalc(w1):
`w1 = z1/z2 ` = w1; ` `;
w2 := r1/r2*exp(I*(t1 - t2)):
`w2 = r1/r2 exp(iPi/2-iPi/3) ` = w2; ` `;
`Does w1 = w2 ?`;
w1 = w2;
evalb(w1 = w2);