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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 z = x+i*y 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 Re(z) = x and Im(z) = y . Geometrically, Re(z) is the projection of z = (x, y) onto the x- axis, and Im(z) is the projection of z 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  z = x+i*y  is a non-negative real number denoted abs(z) and is given by the equation  

    abs(z) = sqrt(x^2+y^2) .    

 The number abs(z) is the distance between the origin and the point (x, y). The only complex number with modulus zero is the number 0. The number z = 4+3*i has modulus 5 . The numbers abs(x) = abs(Re(z)) , abs(y) = abs(Im(z)) , and abs(z) are the lengths of the sides and hypotenuse of a right triangle, from which it follows that

    abs(x) <= abs(z)      and     abs(y) <= abs(z) .

Theorem 1.2  (The triangle inequality)   If z[1] and z[2] are arbitrary comples numbers then

    abs(z[1]+z[2]) <= abs(z[1])+abs(z[2]) .


Example  1.5, Page 19.   Verify the triangle inequality for z[1] and  z[2] .abs(z[1]+z[2]) <= abs(z[1])+abs(z[2])
.

> 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)));

`z1 ` = 7+I

`z2 ` = 3+5*I

`z1 + z2 ` = 10+6*I

` `

`|z1| ` = 5*2^(1/2)

`|z2| ` = 34^(1/2)

`|z1 + z2| ` = 2*34^(1/2)

` `

`|z1 + z2| <= |z1| + |z2|`

2*34^(1/2) <= 5*2^(1/2)+34^(1/2)

11.66190379 <= 12.90201970

true

Example  1.6, Page 20.   Verify that  abs(z[1]*z[2]) = abs(z[1])*abs(z[2]) .

> 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));

z1 := 1+2*I

z2 := 3+2*I

`z1*z2 ` = -1+8*I

` `

`|z1| ` = 5^(1/2)

`|z2| ` = 13^(1/2)

`|z1*z2| ` = 65^(1/2)

` `

`|z1*z2| = |z1|*|z2|`

65^(1/2) = 5^(1/2)*13^(1/2)

65^(1/2) = 65^(1/2)

true

>

End of Section 1.3.