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# Section 1.3 The Geometry of Complex Numbers

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C01-3.mws

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 1.3  The Geometry of Complex Numbers

Since the complex numbers are ordered pairs of real numbers, there is a one-to-one correspondence between them and points in the plane. In this section we shall see what effect algebraic operations on complex numbers have on their geometric representations.

The number
can be represented by a position vector in the xy-plane whose tail is at the origin and whose head is at the point (x,y). When the xy-plane is used for displaying complex numbers, it is called the complex plane, or more simply, the z-plane. Recall that and . Geometrically, is the projection of onto the x- axis, and is the projection of onto the y-axis. It makes sense, then, that the x-axis is also called the real axis, and the y-axis is called the imaginary axis.

Definition 1.8:  Modulus

The modulus, or absolute value, of the complex number    is a non-negative real number denoted and is given by the equation

.

The number is the distance between the origin and the point (x, y). The only complex number with modulus zero is the number 0. The number has modulus . The numbers , , and are the lengths of the sides and hypotenuse of a right triangle, from which it follows that

and      .

Theorem 1.2  (The triangle inequality)   If and are arbitrary comples numbers then

.

Example  1.5, Page 19.
Verify the triangle inequality for and   .
.

 > z1 := 7 + I:  `z1 ` = z1; z2 := 3 + 5*I:  `z2 ` = z2; `z1 + z2 ` = z1 + z2; ` `; `|z1| ` = abs(z1); `|z2| ` = abs(z2); `|z1 + z2| ` = abs(z1 + z2); ` `; `|z1 + z2| <= |z1| + |z2|`; abs(z1 + z2) <= abs(z1) + abs(z2); evalf(abs(z1 + z2) <= abs(z1) + abs(z2)); evalb(evalf(abs(z1 + z2) <= abs(z1) + abs(z2)));

Example  1.6, Page 20.   Verify that   .

 > z1 := 1 + 2*I; z2 := 3 + 2*I; `z1*z2 ` = z1*z2; ` `; `|z1| ` = abs(z1); `|z2| ` = abs(z2); `|z1*z2| ` = abs(z1*z2); ` `; `|z1*z2| = |z1|*|z2|`; abs(z1*z2) = abs(z1)*abs(z2); combine(abs(z1*z2) = abs(z1)*abs(z2),power); evalb(combine(abs(z1*z2) = abs(z1)*abs(z2),power));

 >

End of Section 1.3.