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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) + I*sin(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 := 4*I: `z1 ` = z1; ` `; r  := abs(z1): t := argument(z1): ` r ` = r, theta = t; z2 := r*exp(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);

End of Section 1.4 .

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