COMPLEX ANALYSIS: Maple Worksheets,
2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu
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COMPLEX ANALYSIS: for Mathematics &
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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 ![z[1]](images/C01-484.gif)
= ![r[1]*exp(i*theta[1])](images/C01-485.gif)
and ![z[2]](images/C01-487.gif)
= ![r[2]*exp(i*theta[2])](images/C01-488.gif)
, then as sets
![arg(z[1]*z[2]) = arg(z[1])+arg(z[2])](images/C01-490.gif)
.
Example 1.13, Page 28.
Given ![z[1]](images/C01-491.gif)
and ![z[2]](images/C01-492.gif)
, compute ![z[1]/z[2]](images/C01-493.gif)
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 .