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/

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**

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.