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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 z = x+i*y 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 r be the modulus of z (i.e., r = abs(z) ), and let theta be the angle that the line from the origin to the complex number z makes with the positive x -axis. (Note: The number theta is undefined if z = 0 . Then

(1-25)     z = r(cos(theta)+i*sin(theta)) .

Definition 1.9:  Polar Representation

    The identity  z  =  (r*cos(theta), r*sin(theta) )  =  `r `(cos(theta)+i*sin(theta))  is known as a polar representation of z , and the values r and theta are called polar coordinates of z .

Example 1.7, Page 23.   Find several polar forms of  z = 1+i .

> 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;

`z ` = 1+I

` `

`A few polar forms for z.`

`sqrt(2)cos(Pi/4) + i sqrt(2)sin(Pi/4)` = 1+I

`sqrt(2)cos(9Pi/4) + i sqrt(2)sin(9Pi/4)` = 1+I

`sqrt(2)cos(-7Pi/4) + i sqrt(2)sin(-7Pi/4)` = 1+I

Definition 1.10:  arg(z)

If  z <> 0 , then arg(z) = {theta, where, z = r(cos(theta)+i*sin(theta))} .  

If  theta epsilon arg(z) , we say that theta is an argument of  z .

An argument of  z  is  theta = arg(z)  or  theta = arctan(y/x)  provided that  x <> 0 .
The exponential form of  z  is  z = r*e^(i*theta) ,  where  r = abs(z)  and  theta = arg(z) .  

Example 1.8, Page 24.   Because 1+i = sqrt(2)(cos(pi/4)+i*sin(pi/4)) , we have

    arg(1+i) = {pi/4+2*n*pi, where*n*is*an*integer} .   

Definition 1.11:  Arg(z)

Let z <> 0 be a complex number. Then

Arg(z) = theta , provided z = r(cos(theta)+i*sin(theta)) and -pi <theta <= pi .         

If  theta = Arg(z) , we say that theta is the argument of  z .

Example 1.9, Page 24.  Arg(1+i) = pi/4 .  


Example 1.10, Page 24.  Find the polar form of  z ,  by computing   abs(z)   and  arg(z) .

> 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);

`z1 ` = -3^(1/2)-I

` `

` r ` = 2, theta = -5/6*Pi

`z2 = 2cos(-5Pi/6) + i 2sin(-5Pi/6)` = -3^(1/2)-I

` `

`Does z1 = z2 ?`

-3^(1/2)-I = -3^(1/2)-I

true


Example 1.11, Page 25.   Write  z  in the  z = r*e^(i*theta)  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);

`z1 ` = 4*I

` `

` r ` = 4, theta = 1/2*Pi

`z2 = 4 exp(iPi/2)` = 4*I

` `

`Does z1 = z2 ?`

4*I = 4*I

true


Example 1.12, Page 26.   Given  z ,  find  abs(z)  and  1/z .

> 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;

`z ` = 1+I

`r ` = 2^(1/2)

` `

conjugate(z) = 1-I

` `

conjugate(z)/r^2 = 1/2-1/2*I

`1/z ` = 1/2-1/2*I

Theorem 1.3    If  z[1] = r[1]*exp(i*theta[1]) `` <> 0  and  z[2] = r[2]*exp(i*theta[2]) `` <> 0 , then as sets  

        arg(z[1]*z[2]) = arg(z[1])+arg(z[2]) .     

Example 1.13, Page 28.  Given  z[1]  and  z[2] ,  compute  z[1]/z[2]  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);

`z1 ` = 8*I

`z1 = 8 exp(iPi/2)` = 8*I

` `

`z2 ` = 1+I*3^(1/2)

`z2 = 2 exp(iPi/3)` = 1+I*3^(1/2)

` `

`w1 = z1/z2 ` = 8*I/(1+I*3^(1/2))

`w1 = z1/z2 ` = 2*3^(1/2)+2*I

` `

`w2 = r1/r2 exp(iPi/2-iPi/3) ` = 2*3^(1/2)+2*I

` `

`Does w1 = w2 ?`

2*3^(1/2)+2*I = 2*3^(1/2)+2*I

true

End of Section 1.4 .

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