C07-4.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 7 TAYLOR and LAURENT SERIES

Section 7.4 Singularities, Zeros and Poles

The point alpha is called a singular point , or singularity , of the complex function f if f is not analytic at the point alpha , but every neighborhood D[R](a) of alpha contains at least one point at which f is analytic. For example, the function f(z) = 1/(1-z) is not analytic at alpha = 1 , but is analytic for all other values of z . Thus, the point alpha = 1 is a singular point of f . As another example, consider the function g(z) = Log(z) . We saw in Section 5.2 that g is analytic for all z except at the origin and at the points on the negative real axis. Thus, the origin and each point on the negative real axis is a singularity of g .

The point alpha is called an isolated singularity of a complex function f if f is not analytic at alpha , but there exists a real number `R > 0` such that f is analytic everywhere in the punctured disk D[R]^`*`*`(a)` . The function f(z) = 1/(1-z) has an isolated singularity at alpha = 1 .

The function g(z) = Log(z) , however, has a singularity at alpha = 0 (or at any point of the negative real axis) that is not isolated, because any neighborhood of alpha will contain points on the negative real axis, and g is not analytic at those points. Functions with isolated singularities have a Laurent series because the punctured disk D[R]^`*`*`(a)` is the same as the annulus A(alpha,0,R) . We now look at this special case of Laurent's theorem in order to classify three types of isolated singularities.

Example for a pole, Page 291. Consider the function f(z) = cot(z) .
The leading term in the series expansion S(z) is 1/z and
 limit(S(z),z = 0) = infinity in the same manner as limit(cot(z),z = 0) = infinity .

> f:='f': L:='L': s:='s': S:='S': z:='z': Z:='Z':
f := z -> cot(z):
S := series(f(Z), Z=0, 8):
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
`f(z) ` = f(z);
`f(z) ` = subs(Z=z,S);
L[7](z) = LS(z);

`f(z) ` = cot(z)

`f(z) ` = series(1*z^(-1)-1/3*z-1/45*z^3-2/945*z^5-1/4725*z^7+O(z^8),z,8)

L[7](z) = 1/z-1/3*z-1/45*z^3-2/945*z^5-1/4725*z^7

> x:='x': y:='y':
plot({f(x),LS(x)}, x=-3.14..3.14, y=-15..15,
title=`y = cot(x) and y = L7(x)`,
tickmarks=[7,7]);

[Maple Plot]

Example of a removable singularity, Page 291. The function f(z) = sin(z)/z
has a removable singularity at z = 0 .

> f:='f': L:='L': s:='s': S:='S': z:='z': Z:='Z':
f := z -> sin(z)/z:
`f(z) ` = f(z);
S := series(sin(Z), Z=0, 11)/Z:
`f(z) ` = subs(Z=z,S);
S := series(f(Z), Z=0, 11):
`f(z) ` = subs(Z=z,S);
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
L[8](z) = LS(z);

`f(z) ` = sin(z)/z

`f(z) ` = (series(1*z-1/6*z^3+1/120*z^5-1/5040*z^7+1/362880*z^9+O(z^11),z,11))/z

`f(z) ` = series(1-1/6*z^2+1/120*z^4-1/5040*z^6+1/362880*z^8+O(z^10),z,10)

L[8](z) = 1-1/6*z^2+1/120*z^4-1/5040*z^6+1/362880*z^8


Example of a removable singularity, Page 291. The function f(z) = (cos(z)-1)/(z^2)
has a removable singularity at z = 0 .

> f:='f': L:='L': s:='s': S:='S': z:='z': Z:='Z':
f := z -> (cos(z) -1)/z^2:
`f(z) ` = f(z);
S := series((cos(Z) -1), Z=0, 11)/Z^2:
`f(z) ` = subs(Z=z,S);
S := series(f(Z), Z=0, 11):
`f(z) ` = subs(Z=z,S);
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
`f(z) ` = f(z);
L[8](z) = LS(z);

`f(z) ` = (cos(z)-1)/z^2

`f(z) ` = (series(-1/2*z^2+1/24*z^4-1/720*z^6+1/40320*z^8-1/3628800*z^10+O(z^11),z,11))/z^2

`f(z) ` = series(-1/2+1/24*z^2-1/720*z^4+1/40320*z^6-1/3628800*z^8+O(z^9),z,9)

`f(z) ` = (cos(z)-1)/z^2

L[8](z) = -1/2+1/24*z^2-1/720*z^4+1/40320*z^6-1/3628800*z^8

Example of a pole of order 2, Page 291. The function f(z) = sin(z)/(z^3)
has a pole of order n = 2 at z = 0 .

> f:='f': L:='L': s:='s': S:='S': z:='z': Z:='Z':
f := z -> sin(z)/z^3:
`f(z) ` = f(z);
S := series(sin(Z), Z=0, 13)/Z^3:
`f(z) ` = subs(Z=z,S);
S := series(f(Z), Z=0, 13):
`f(z) ` = subs(Z=z,S);
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
L[8](z) = LS(z);

`f(z) ` = sin(z)/z^3

`f(z) ` = (series(1*z-1/6*z^3+1/120*z^5-1/5040*z^7+1/362880*z^9-1/39916800*z^11+O(z^13),z,13))/z^3

`f(z) ` = series(1*z^(-2)-1/6+1/120*z^2-1/5040*z^4+1/362880*z^6-1/39916800*z^8+O(z^10),z,10)

L[8](z) = 1/(z^2)-1/6+1/120*z^2-1/5040*z^4+1/362880*z^6-1/39916800*z^8

Example of a simple pole, Page 291. The function f(z) = exp(z)/z
has a simple pole at z = 0 .

> f:='f': L:='L': s:='s': S:='S': z:='z': Z:='Z':
f := z -> exp(z)/z:
`f(z) ` = f(z);
S := series(exp(Z), Z=0, 8)/Z:
`f(z) ` = subs(Z=z,S);
S := series(f(Z), Z=0, 8):
`f(z) ` = subs(Z=z,S);
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
L[6](z) = LS(z);

`f(z) ` = exp(z)/z

`f(z) ` = (series(1+1*z+1/2*z^2+1/6*z^3+1/24*z^4+1/120*z^5+1/720*z^6+1/5040*z^7+O(z^8),z,8))/z

`f(z) ` = series(1*z^(-1)+1+1/2*z+1/6*z^2+1/24*z^3+1/120*z^4+1/720*z^5+1/5040*z^6+O(z^7),z,7)

L[6](z) = 1/z+1+1/2*z+1/6*z^2+1/24*z^3+1/120*z^4+1/720*z^5+1/5040*z^6


Examples of an essential singularity, Page 291. The function f(z) = z^2*sin(1/z)
has an essential singularity at z = 0 .

> f:='f': L:='L': p:='p': s:='s': z:='z': Z:='Z':
f := z -> z^2*sin(1/z):
s := convert(series(sin(z), z=0, 12), polynom):
S := subs(z=1/Z,s):
p := z -> subs(Z=z, expand(Z^2 * S)):
`f(z) ` = f(z);
L[9](z) = p(z);

`f(z) ` = z^2*sin(1/z)

L[9](z) = z-1/6*1/z+1/120/z^3-1/5040*1/(z^5)+1/362880/z^7-1/39916800*1/(z^9)

There will be infinitely many terms involving negative powers of z ,
hence, f(z) has an essential singularity at z = 0 .

Theorem 7.10 A function f(z) analytic in D[R](a) has a zero of order k
at the point z = alpha if and only if its Taylor series given by f(z) = sum(c[n]*(z-alpha)^n,n = 0 .. infinity)
has c[0] = 0, c[1] = 0, `...`, c[k-1] = 0 and c[k] <> 0 .


Example 7.10, Page 292. Show that f(z) = z*sin(z^2) has a zero of order 3 at z = 0 .

> f:='f': p:='p': P:='P': s:='s': z:='z': Z:='Z':
f := z -> z*sin(z^2):
s := convert(series(f(Z), Z=0, 24), polynom):
p := z -> subs(Z=z,s):
`f(z) ` = f(z);
P[23](z) = p(z);

`f(z) ` = z*sin(z^2)

P[23](z) = z^3-1/6*z^7+1/120*z^11-1/5040*z^15+1/362880*z^19-1/39916800*z^23

Thus, f(z) = z*sin(z^2) has a zero of order 3 at z = 0 .

Theorem 7.11 Suppose f(z) is analytic in D[R](a) . Then f(z) has a zero of order k at the point z = alpha if and only if it can be expressed in the form f(z) = (z-alpha)^k*g(z) , where g(z) is analytic at z = alpha and g(alpha) <> 0 .

Corollary 7.4 If f(z) and g(z) are analytic at z = alpha and have zeros of orders m and n , respectively,

then their product h(z) = f(z)*g(z) has a zero of order m+n at the point z = alpha .


Example 7.11, Page 293. Show that f(z) = z^3*sin(z) has a zero of order 4 at z = 0 .

> f:='f': p:='p': P:='P': s:='s': z:='z': Z:='Z':
f := z -> z^3*sin(z):
s := convert(series(f(Z), Z=0, 15), polynom):
p := z -> subs(Z=z,s):
`f(z) ` = f(z);
P[14](z) = p(z);

`f(z) ` = z^3*sin(z)

P[14](z) = z^4-1/6*z^6+1/120*z^8-1/5040*z^10+1/362880*z^12-1/39916800*z^14

Thus, f(z) = z^3*sin(z) has a zero of order 4 at z = 0 .

Theorem 7.12 A function f(z) analytic in the punctured disk D[R](a)^`*` has a pole of order k at z = alpha if and only if it can be expressed in the form

f(z) = h(z)/((z-alpha)^k) ,

where h(z) is analytic at z = alpha and h(alpha) <> 0 .

Corollary 7.5 If f(z) is analytic and has a zero of order k at z = alpha ,

then g(z) = 1/f(z) has a pole of order k at z = alpha .

Corollary 7.6 If f(z) has a pole of order k at z = alpha , then h(z) = 1/f(z) has a removable singularity at z = alpha .

If we define h(alpha) = 0 , then h(z) has a zero of order k at z = alpha .


Example 7.12, Page 295. Locate the zeros and poles of h(z) = tan(z)/z ,
and determine their order.

> h:='h': L:='L': s:='s': s:='s': z:='z': Z:='Z':
h := z -> tan(z)/z:
`h(z) ` = h(z);
S := series(h(Z), Z=0, 12):
`s(z) ` = subs(Z=z,S);
s := convert(S, polynom):
LS := z -> subs(Z=z,s):
L[10](z) = LS(z);

`h(z) ` = tan(z)/z

`s(z) ` = series(1+1/3*z^2+2/15*z^4+17/315*z^6+62/2835*z^8+1382/155925*z^10+O(z^11),z,11)

L[10](z) = 1+1/3*z^2+2/15*z^4+17/315*z^6+62/2835*z^8+1382/155925*z^10

Thus, h(z) = tan(z)/z has a removable singularity at z = 0 .

Next consider the points z = ... , 2*pi, -pi, pi, 2*pi , ... .

> tan(pi)/pi = tan(Pi)/Pi;
tan(2*pi)/(2*pi) = tan(2*Pi)/(2*Pi);
tan(3*pi)/(3*pi) = tan(3*Pi)/(3*Pi);
tan(4*pi)/(4*pi) = tan(4*Pi)/(4*Pi);
tan(5*pi)/(5*pi) = tan(5*Pi)/(5*Pi);

tan(pi)/pi = 0

1/2*tan(2*pi)/pi = 0

1/3*tan(3*pi)/pi = 0

1/4*tan(4*pi)/pi = 0

1/5*tan(5*pi)/pi = 0

Thus h(z) = tan(z)/z has simple zeros at z = n*pi where `n = ... -2, -1, 1, 2, ...` .

Next consider the points z = ..., pi/2-2*pi, pi/2-pi, pi/2, pi/2+pi, pi/2+2*pi , ... .

> tan(pi/2)/(pi/2) = limit(h(x),x=Pi/2,left);
tan(3*pi/2)/(3*pi/2) = limit(h(x),x=3*Pi/2,left);
tan(5*pi/2)/(5*pi/2) = limit(h(x),x=5*Pi/2,left);
tan(7*pi/2)/(7*pi/2) = limit(h(x),x=7*Pi/2,left);

2*tan(1/2*pi)/pi = infinity

2/3*tan(3/2*pi)/pi = infinity

2/5*tan(5/2*pi)/pi = infinity

2/7*tan(7/2*pi)/pi = infinity

Thus h(z) = tan(z)/z has poles at z = ..., pi/2-2*pi, pi/2-pi, pi/2, pi/2+pi, pi/2+2*pi , ... .

> s := z -> series(h(z), z=Pi/2, 3):
`h(z) ` = h(z);
`h(z) ` = s(z);

`h(z) ` = tan(z)/z

`h(z) ` = series((-2*1/Pi)*(z-1/2*Pi)^(-1)+4*1/(Pi^2)+2*(1/3-4/Pi^2)/Pi*(z-1/2*Pi)+(-4/3*(Pi^2-12)/Pi^4)*(z-1/2*Pi)^2+O((z-1/2*Pi)^3),z=-(-1/2*Pi),3)

Hence h(z) = tan(z)/z has simple poles at z = (n+1/2)*pi where n is an integer.


Example 7.13, Page 296. Locate the poles of g(z) = 1/(5*z^4+26*z^2+5) ,
and specify their order.

> g:='g': z:='z':
g := z -> 1/(5*z^4 + 26*z^2 + 5):
`g(z) ` = g(z);

`g(z) ` = 1/(5*z^4+26*z^2+5)

Find the singularities of g(z) = 1/(5*z^4+26*z^2+5) .

> Zn := sort([solve(denom(g(z))=0, z)]):
`For g(z) ` = g(z);
`The singularities are:`;
z1 := subs(z=Zn[1],z): z[1] = z1;
z2 := subs(z=Zn[2],z): z[2] = z2;
z3 := subs(z=Zn[3],z): z[3] = z3;
z4 := subs(z=Zn[4],z): z[4] = z4;

`For g(z) ` = 1/(5*z^4+26*z^2+5)

`The singularities are:`

z[1] = I*sqrt(5)

z[2] = -I*sqrt(5)

z[3] = 1/5*I*sqrt(5)

z[4] = -1/5*I*sqrt(5)

Thus g(z) = 1/(5*z^4+26*z^2+5) , has four simple poles.


Example 7.14, Page 296. Locate the zeros and poles of g(z) = pi*cot(pi*z)/(z^2) ,
and determine their order.

> g:='g': L:='L': p:='p': s:='s': S:='S': z:='z': Z:='Z':
g := z -> Pi*cot(Pi*z)/z^2:
s := series(Pi*cot(Pi*z), z=0, 7)/z^2:
S := series(g(z), z=0, 7):
p := convert(series(Pi*cot(Pi*Z), Z=0, 7), polynom):
LS := z -> subs(Z=z,expand(p/z^2)):
`g(z) ` = g(z);
`g(z) ` = s;
`g(z) ` = S;
L[3](z) = LS(z);

`g(z) ` = Pi*cot(Pi*z)/z^2

`g(z) ` = (series(1*z^(-1)+(-1/3*Pi^2)*z+(-1/45*Pi^4)*z^3+(-2/945*Pi^6)*z^5+O(z^7),z,7))/z^2

`g(z) ` = series(1*z^(-3)+(-1/3*Pi^2)*z^(-1)+(-1/45*Pi^4)*z+(-2/945*Pi^6)*z^3+O(z^5),z,5)

L[3](z) = 1/(z^3)-1/3*1/z*Pi^2-1/45*z*Pi^4-2/945*z^3*Pi^6

Thus g(z) = pi*cot(pi*z)/(z^2) , has a pole of order n = 3 at z = 0 .

Next consider the points z = ... , -3, -2, -1, 1, 2, 3 , ... .

> pi*cot(pi)/(pi)^2 = limit(g(x),x=1,right);
2*pi*cot(2*pi)/(2*pi)^2 = limit(g(x),x=2,right);
3*pi*cot(3*pi)/(3*pi)^2 = limit(g(x),x=3,right);
s := z -> series(g(z), z=1, 5):
`g(z) ` = g(z),` Expand about the point `,z[0] = 1;
`g(z) ` = s(z);
s := z -> series(g(z), z=2, 5):
`g(z) ` = g(z),` Expand about the point `,z[0] = 2;
`g(z) ` = s(z);

1/pi*cot(pi) = infinity

1/2/pi*cot(2*pi) = infinity

1/3/pi*cot(3*pi) = infinity

`g(z) ` = Pi*cot(Pi*z)/z^2, ` Expand about the point `, z[0] = 1

`g(z) ` = series(1*(z-1)^(-1)-2+(3-1/3*Pi^2)*(z-1)+(-4+2/3*Pi^2)*(z-1)^2+(5-Pi^2-1/45*Pi^4)*(z-1)^3+O((z-1)^4),z=-(-1),4)

`g(z) ` = Pi*cot(Pi*z)/z^2, ` Expand about the point `, z[0] = 2

`g(z) ` = series(1/4*(z-2)^(-1)-1/4+(3/16-1/12*Pi^2)*(z-2)+(-1/8+1/12*Pi^2)*(z-2)^2+(5/64-1/16*Pi^2-1/180*Pi^4)*(z-2)^3+O((z-2)^4),z=-(-2),4)

Thus g(z) = pi*cot(pi*z)/(z^2) has simple poles at the points z = ... , -3, -2, -1, 1, 2, 3 , ... . .

End of Section 7.4.