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 5 ELEMENTARY FUNCTIONS

Section 5.3 Complex Exponents

In Section 1.5 we indicated that the complex numbers are complete in the sense that it is possible to make sense out of expressions such as sqrt(1+i) or (-1)^i left without appealing to a number system beyond the framework of complex numbers. We will do this by taking note of some rudimentary properties of the complex exponential and logarithm, and then using our imagination.

 

 

 

Definition 5.4: Complex exponent

 

Let c be a complex number. We define z^c as

 

z^c = exp( c*log(z) ) .

 


 Example 5.7, Page 178. (a) Find the principal value of sqrt(1+i) .

> c:='c': C:='C': w:='w': z:='z': Z:='Z':
Z := 1 + I:
C := 1/2:
z = Z;
c = C;
w := exp(C*log(Z)):
z^c = w;
w := evalc(w):
z^c = w;
w := exp(C*evalc(log(Z))):
z^c = w;
w := evalf(w):
z^c = w;

z = 1+I

c = 1/2

z^c = sqrt(1+I)

z^c = 1/2*sqrt(2+2*sqrt(2))+1/2*I*sqrt(-2+2*sqrt(2)...

z^c = exp(1/4*ln(2)+1/8*I*Pi)

z^c = 1.098684113+.4550898607*I

 

(b) Find the principal value of (-1)^i .

> c:='c': C:='C': w:='w': z:='z': Z:='Z':
Z := -1:
C := I:
w := Z^C:
z^c = w;
w := exp(C*log(Z)):
z^c = w;
w := evalf(w):
z^c = w;

z^c = (-1)^I

z^c = exp(-Pi)

z^c = .4321391825e-1

 

Example 5.8, Page 179. Find the values of 2^(1/9+i/50) .
The principal value is:

> c:='c': C:='C': w:='w': z:='z': Z:='Z':
Z := 2:
C := 1/9 + I/50:
w := Z^C:
z^c = w;
w := exp(C*log(Z)):
z^c = w;
w := evalc(w):
z^c = w;
w := evalf(w):
z^c = w;

z^c = 2^(1/9+1/50*I)

z^c = exp((1/9+1/50*I)*ln(2))

z^c = 2^(1/9)*cos(1/50*ln(2))+I*2^(1/9)*sin(1/50*ln...

z^c = 1.079955957+.1497232768e-1*I

 

Some of the other values of 2^(1/9+i/50) are:

> n:='n': z:='z': Z:='Z':
Z := n -> evalf(exp((1/9+I/50)*(log(2)+2*Pi*I*n))):
z[n] = exp((1/9+I/50)*(log(2)+2*Pi*I*n)); ` `;
for n from -5 to 5 do print(Z(n)) od;

z[n] = exp((1/9+1/50*I)*(ln(2)+2*I*Pi*n))

` `

-1.911846766+.6659891963*I

-1.669148859-.6338597718*I

-.7683256028-1.374435038*I

.2600741246-1.364095102*I

.9489822980-.7741286217*I

1.079955957+.1497232768e-1*I

.7211126821+.6223223308*I

.1343885474+.8292161686*I

-.3792768168+.6363870140*I

-.6169892442+.2149273564*I

-.5386660228-.2045585810*I

> pts := [[ Re(Z(j)), Im(Z(j))] $j=-9..9]:
plot(pts, x=-2..3,
title=`Values of 2^(1/9 + I/50).`,
scaling=constrained,
labels=[` x`,`y `],
style=point,symbol=circle,
tickmarks=[5,4],
view=[-2.0..2.6,-1.5..2.6]);

[Maple Plot]

 

Example 5.9, Page 180. (a) Find the values of (i^2)^i . The principal values is:

> c:='c': C:='C': w:='w': z:='z': Z:='Z':
Z := I^2:
C := I:
w := Z^C:
z^c = w;
w := exp(C*log(Z)):
z^c = w;
w := evalf(w):
z^c = w;

z^c = (-1)^I

z^c = exp(-Pi)

z^c = .4321391825e-1

 

(b) Find the values of i^(2*i) .
The principal values is:

> c:='c': C:='C': w:='w': z:='z': Z:='Z':
Z := I:
C := 2*I:
w := Z^C:
z^c = w;
w := exp(C*log(Z)):
z^c = w;
w := evalf(w):
z^c = w;

z^c = I^(2*I)

z^c = exp(-Pi)

z^c = .4321391825e-1

 

Remark. The principal value is the same, but the general values are different.
Some of the other values are given in the sets:

> Z1 := n -> evalf(exp(I*(log(I^2)+2*Pi*I*n))):
Z2 := n -> evalf(exp(2*I*(log(I)+2*Pi*I*n))):
`Portions of the solution sets are:`;
Z1s := {[ Re(Z1(j)), Im(Z1(j))] $j=-2..2}:
`(i^2)^i ` = Z1s;
Z2s := {[ Re(Z2(j)), Im(Z2(j))] $j=-2..2}:
`i^(2i) ` = Z2s;

`Portions of the solution sets are:`

`(i^2)^i ` = {[12391.64787, 0.], [23.14069264, 0.],...

`i^(2i) ` = {[3553321298., 0.], [12391.64782, 0.], ...
`i^(2i) ` = {[3553321298., 0.], [12391.64782, 0.], ...

Remark. The two sets are different. Therefore, in general, (z^a)^b <> z^ab .

 

End of Section 5.3.