C07-5.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 7.5 Applications of Taylor and Laurent Series

In this section we show how Taylor and Laurent series can be used to derive important properties of analytic functions. We begin by showing that the zeros of an analytic function must be "isolated" unless the function is identically zero. A point alpha of a set T is called isolated if there exists a disk D[r](alpha) about alpha that does not contain any other points of T .

Theorem 7.13 Suppose f(z) is analytic at z = alpha and that f(alpha) = 0 . If f(z) is
not identically zero, then there exists a punctured disk
D[r](alpha)^`*` in which f(z) has no zeros.

Corollary 7.11 (L'Hopital's rule) Suppose f(z) and g(z) are analytic at z = alpha .


If
f(alpha) = 0 and g(alpha) = 0 but `g '`(alpha) <> 0 , then limit(f(z)/g(z),z = alpha) = limit(`f '`(z)/`g '`(z),z = alpha) = `f '`(alpha)/`g '`(alpha) .

Theorem 7.14 (Division of power series)
Suppose
f(z) and g(z) are analytic at z = alpha with power series representations
f(z) = sum(a[n]*(z-alpha)^n,n = 0 .. infinity) for z*epsilon*D[R](alpha) and g(z) = sum(b[n]*(z-alpha)^n,n = 0 .. infinity) for z*epsilon*D[R](alpha) .
If
g(alpha) <> 0 , then the quotient f(z)/g(z) has the power series representation
f(z)/g(z) = sum(c[n]*(z-alpha)^n,n = 0 .. infinity) , where the coefficients satisfy the equations
a[n] = b[0]*c[n]+b[1]*c[n-1]+`...`+b[n-1]*c[1]+b[n]*c[0] .
In other words, the series for the quotient
f(z)/g(z) can be obtained by the familiar process of
dividing the series for
f(z) by the series for g(z) using the standard long division algorithm.


Example 7.15, Page 300. Find the firt few terms of the
Maclaurin series for
f(z) = sec(z) and use it to compute f^` (4)`*`(0)` .

> f:='f': p:='p': P:='P': S:='S': z:='z': Z:='Z':
f := z -> sec(z):
S := series(f(Z), Z=0, 14):
p := z -> subs(Z=z,convert(S, polynom)):
`f(z) ` = f(z);
`f(z) ` = subs(Z=z,S);
P[12](z) = p(z);

`f(z) ` = sec(z)

`f(z) ` = series(1+1/2*z^2+5/24*z^4+61/720*z^6+277/8064*z^8+50521/3628800*z^10+540553/95800320*z^12+O(z^14),z,14)

P[12](z) = 1+1/2*z^2+5/24*z^4+61/720*z^6+277/8064*z^8+50521/3628800*z^10+540553/95800320*z^12

Or we could use Maple's "unapply" procedure.

> f:='f': p:='p': P:='P': s:='s': t:='t': z:='z': Z:='Z':
f := z -> sec(z):
S := taylor(f(z), z=0, 14):
p:=unapply(convert(taylor(f(z),z=0,14),polynom),z):
`f(z) ` = f(z);
`f(z) ` = S;
P[12](z) = p(z);

`f(z)  ` = sec(z)

`f(z)  ` = series(1+1/2*z^2+5/24*z^4+61/720*z^6+277/8064*z^8+50521/3628800*z^10+540553/95800320*z^12+O(z^14),z,14)

P[12](z) = 1+1/2*z^2+5/24*z^4+61/720*z^6+277/8064*z^8+50521/3628800*z^10+540553/95800320*z^12

> D4 := simplify(diff(f(Z), Z$4)):
f4 := z -> subs(Z=z, D4):
S4 := simplify(diff(p(Z), Z$4)):
p4 := z -> subs(Z=z, S4):
`f'(z) ` = simplify(diff(f(z), z));
`p'(z) ` = simplify(diff(p(z), z)); ` `;
`f''(z) ` = simplify(diff(f(z), z$2));
`p''(z) ` = simplify(diff(p(z), z$2)); ` `;
`f'''(z) ` = simplify(diff(f(z), z$3));
`p'''(z) ` = simplify(diff(p(z), z$3)); ` `;
`f''''(z) ` = f4(z);
`p''''(z) ` = p4(z); ` `;
`f''''(0) ` = eval(f4(0));
`p''''(0) ` = p4(0);

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

`p'(z) ` = z+5/6*z^3+61/120*z^5+277/1008*z^7+50521/362880*z^9+540553/7983360*z^11

` `

`f''(z) ` = -(cos(z)^2-2)/cos(z)^3

`p''(z) ` = 1+5/2*z^2+61/24*z^4+277/144*z^6+50521/40320*z^8+540553/725760*z^10

` `

`f'''(z) ` = -sin(z)*(cos(z)^2-6)/cos(z)^4

`p'''(z) ` = 5*z+61/6*z^3+277/24*z^5+50521/5040*z^7+540553/72576*z^9

` `

`f''''(z) ` = (cos(z)^4-20*cos(z)^2+24)/cos(z)^5

`p''''(z) ` = 5+61/2*z^2+1385/24*z^4+50521/720*z^6+540553/8064*z^8

` `

`f''''(0) ` = 5

`p''''(0) ` = 5

Theorem 7.15 (Riemann) Suppose that f(z) is analytic in D[r](alpha)^`*` .
If
f(z) is bounded in D[r](alpha)^`*` , then either f(z) is analytic at z = alpha or f(z) has a removable
singularity at
z = alpha .

Theorem 7.16 The function f(z) has a pole of order k at z = alpha
if and only if
limit(abs(f(z)),z = alpha) = infinity .

Theorem 7.18 The function f(z) has an essential singularity at z = alpha
if and only if
limit(abs(f(z)),z = alpha) does not exist.

Example 7.16, Page 302. Show that the function g(z) = exp(-1/(z^2))
is NOT continuous at
z = 0 .

> g:='g': G:='G': x:='x': y:='y': z:='z':
G := z -> exp(-1/z^2):
`g(z) ` = G(z); ` `;
`g(x + I*0) ` = evalc(G(x + I*0)),` and `,
`g(0 + I*y) ` = evalc(G(0 + I*y));
Limit(g(x), x=0, right),` = `,
Limit(G(x), x=0, right) = limit(G(x), x=0, right);
Limit(g(0 + I*y), y=0, right),` = `,
Limit(G(I*y), y=0, right) = limit(G(I*y), y=0, right);

`g(z) ` = exp(-1/(z^2))

` `

`g(x + I*0) ` = exp(-1/(x^2)), `  and  `, `g(0 + I*y) ` = exp(1/(y^2))

Limit(g(x),x = 0,right), ` = `, Limit(exp(-1/(x^2)),x = 0,right) = 0

Limit(g(I*y),y = 0,right), ` = `, Limit(exp(1/(y^2)),y = 0,right) = infinity

Since the limits are NOT the same, g(z) = exp(-1/(z^2)) is NOT continuous at z = 0 .
REMARK. Although
g(z) = exp(-1/(z^2)) is known to be infinitely differentiable,
it does not have a Maclaurin series expansion. However, it does have a Laurent series.

> f:='f': p:='p': S:='S': z:='z':
f := z -> exp(-1/z^2):
s := z -> series(f(z), z=infinity, 14):
`f(z) ` = f(z);
`f(z) ` = s(z);

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

`f(z) ` = 1-1/(z^2)+1/2/z^4-1/6*1/(z^6)+1/24/z^8-1/120*1/(z^10)+1/720/z^12+O(1/(z^14))

End of Section 7.5.