COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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CHAPTER 7 TAYLOR and LAURENT SERIES

Section 7.3 Laurent series Representations

Suppose is not analytic in but is analytic in the punctured disk ={ }. For example, the function is not analytic when but is analytic for . Clearly, this function does not have a Maclaurin series representation. If we use the Maclaurin series for however, and formally divide each term in that series by , we obtain the representation

= = + + + + + + +

which is valid for all such that . This example raises the question as to whether it might be possible to generalize the Taylor series method to functions analytic in an annulus

= { }.

Perhaps we can represent these functions with a series that employs negative powers of in some way as we did with . As you will see shortly, this is indeed the case. We begin by defining a series that allows for negative powers of .

Theorem 7.7 (Laurent series)

Suppose the Laurent series converges on an annulus .

Then the series converges uniformly on any subannulus where .

Example on page 280. Find the Laurent series for .

> f:='f': z:='z':
f := z -> exp(z)/z^3:
`f(z) ` = f(z);

The following Laurent series is valid for .

> L:='L': s:='s': S:='S': z:='z': Z:='Z':
s := taylor(exp(z), z=0, 9)/z^3:
`f(z) ` = exp(z)/z^3;
`f(z) ` = s;
s := series(exp(z)/z^3, z=0, 9):
`f(z) ` = s;
S := series(exp(Z)/Z^3, Z=0, 9):
S := convert(S, polynom):
LS := z -> subs(Z=z,S):
L[5](z) = LS(z);

Sum up the terms to verify that we have things right.

> L5 := z -> sum(z^(n-3)/n!, n=0..8):
L0 := z -> sum(z^(n-3)/n!, n=0..infinity):
`f(z) ` = exp(z)/z^3;
L[5](z),` = `,Sum(z^(n-3)/n!, n=0..8) = L5(z);
L[infinity](z),` = `,Sum(z^(n-3)/n!, n=0..infinity) = L0(z);

We can plot the real functions and
to get the idea that is an approximation to .

> plot({f(x),LS(x)},
x=0.01..8, y=0..25,
title=`y=exp(x)/x^3 and y=L5(x)`,
labels=[` x`,`y `],
tickmarks=[8,4],
view=[0..8,0..25]);

Theorem 7.8 (Laurent's theorem) Suppose , and that is analytic in the annulus .

If is any number such that , then for all , has the Laurent series representation

= = ,

where for the coefficients and are given by

=

and

= .

Theorem 7.9 (Uniqueness and differentiation of Larent series)

Suppose that is analytic in the annulus , and has the Laurent series representation

for all .
(i) If for all , then for all n.
(In other words, the Laurent series for in a given annulus is unique.)
(ii) For all , the derivatives for may be obtained by termwise
differentiation of its Laurent series.

Example 7.7, Page 286. Find three different Laurent series representations for
involving powers of .

> f:='f': z:='z':
f := z -> 3/(1+z)/(2-z):
`f(z)` = f(z);

Maple easily finds the Maclaurin series involving positive powers of :

> L:='L': Z:='Z':
s0 := z -> subs(Z=z, series(f(Z), Z=0, 6)):
`f(z) ` = f(z);
`f(z) ` = s0(z);
S0 := convert(s0(Z), polynom):
L0 := z -> subs(Z=z,S0):
L[0](z) = L0(z);

The above series converges for , this requires looking at the sum of the two"geometric series"
that form the parts of . Maple can also find "the other series" that involves negative powers of ,
this involves the mental thinking that you substitute in the original series to get a new function of ,
then expand it about and then substitute .

> Z:='Z':
Se := series(f(1/Z), Z=0, 7):
s1 := z -> subs(Z=1/z,Se):
`f(z) ` = f(z);
`f(z) ` = s1(z);
S1 := convert(series(f(1/Z), Z=0, 7), polynom):
S1 := subs(Z=1/Z,S1):
L1 := z -> subs(Z=z,S1):
L[1](z) = L1(z);

Or we can use Maple and expand about .

> `f(z) ` = f(z);
`f(z) ` = series(f(z), z=infinity, 7);

The above series converges for , this requires looking at the sum of the two "geometric series"
that form the parts of . This leaves an unresolved question: "What happens when .
This requires a little work. First split up the functions up into their partial fraction form, and then make
expansions about and for each part.

> F:='F': f1:='f1': f2:='f2':
A := convert(f(z), parfrac, z): `f(z) ` = A;
f1 := z -> 1/(1+z): F[1](z) = f1(z);
f2 := z -> -1/(z-2): F[2](z) = f2(z);

Now form Laurent expansions for the two parts and of .

> S:='S': Z:='Z':
S11 := series(f1(z), z=0, 20):
S11 := convert(S11, polynom):
S[11](z) = S11;
S12 := series(f1(z), z=infinity, 18):
S12 := series(f1(1/Z), Z=0, 18):
S12 := convert(S12, polynom):
S12 := subs(Z=1/z,S12):
S[12](z) = S12;
S21 := series(f2(z), z=0, 12):
S21 := convert(S21, polynom):
S[21](z) = S21;
S22 := series(f2(z), z=infinity, 12):
S22 := series(f2(1/Z), Z=0, 12):
S22 := convert(S22, polynom):
S22 := subs(Z=1/z,S22):
S[22](z) = S22;

The Laurent series is valid for .

> Z:='Z':
s1 := subs(z=Z,S11+S21):
L1 := z -> subs(Z=z,s1):
`f(z) ` = f(z);
L[1](z) = L1(z);

> plot({f(x),L1(x)},x=0.0..1,
title=`s = f(x), s = L1(x)`,
tickmarks=[5,3],
view=[0..1,1.3..1.5]);

The following Laurent series is valid for .

> Z:='Z':
s2 := subs(z=Z,S12+S21):
L2 := z -> subs(Z=z,s2):
`f(z) ` = f(z);
L[2](z) = L2(z);

> plot({f(x),L2(x)},x=1.01..1.99,
title=`s = f(x), s = L2(x)`,
tickmarks=[5,3],
view=[1..2,1.5..4]);

So is an approximation to valid for . Notice that is
NOT a good approximation for , for example it is not accurate at .

> `f(1.5) L1(1.5)`;
f(1.5) <> L1(1.5); ` `;
`f(1.5) ~ L2(1.5)`;
f(1.5), `~` , L2(1.5);

The Laurent series is valid for .

> Z:='Z':
s3 := subs(z=Z,S12+S22):
L3 := z -> subs(Z=z,s3):
`f(z) ` = f(z);
L[3](z) = L3(z);

> plot({f(x),L3(x)},x=2.01..4.0,
title=`s = f(x), s = L3(x)`,
tickmarks=[2,6],
view=[2..4,-5..0]);

So is an approximation to valid for . Notice that and
are NOT good approximations for , for example they are not accurate at .

> `f(4.0) L1(4.0)`;
f(4.0) <> L1(4.0);` `;
`f(4.0) L2(4.0)`;
f(4.0) <> L2(4.0);` `;
`f(4.0) ~ L3(4.0)`;
f(4.0), `~` , L3(4.0);

The fourth possibility does not converge for any values of z and should not be considered,
however you could write down a few terms and think about why this series diverges.

> Z:='Z':
s4 := subs(z=Z,S11+S22):
L4 := z -> subs(Z=z,s4):
L[4](z) = L4(z);

In the above series, when the portion of the series with positive exponents will converge
but the portion with negative exponents will diverge. Similarly, when the portion of the series
with negative exponents will converge, but the portion of the series with positive exponents will diverge.
Hence, the series diverges for all .

> `f(0.5) L4(0.5)`;
f(0.5) <> L4(0.5);` `;
`f(1.5) L4(1.5)`;
f(1.5) <> L4(1.5);` `;
`f(4.0) L4(4.0)`;
f(4.0) <> L4(4.0);

Example 7.8, Page 286. Find the Laurent series for .

> f:='f': z:='z':
f := z ->(cos(z) - 1)/z^4:
`f(z) ` = f(z);

The following Laurent series is valid for .

> L:='L': s:='s': S:='S': Z:='Z':
S := series(cos(Z)-1, Z=0, 12)/Z^4:
s := convert(series(cos(Z)-1, Z=0, 12), polynom)/Z^4:
LS := z -> subs(Z=z,expand(s)):
`f(z) ` = f(z);
`f(z) ` = subs(Z=z,S);
`f(z) ` = series((cos(z) - 1)/z^4, z=0, 12);
L[6](z) = LS(z);

Sum up the terms to verify that we have things right.

> L8 := z -> sum((-1)^n * z^(2*n-4)/(2*n)!, n=1..6):
L0 := z -> sum((-1)^n * z^(2*n-4)/(2*n)!, n=1..infinity):
`f(z) ` = (cos(z) - 1)/z^4;
L[8](z),` = `,
Sum((-1)^n * z^(2*n-4)/(2*n)!, n=1..6) = L8(z);
L[infinity](z),` = `,
Sum((-1)^n * z^(2*n-4)/(2*n)!, n=1..infinity) = L0(z);

Compare the graph of the function and its Laurent series.

> plot({f(x),LS(x)}, x=0.01..5.0,
title=`y = f(x), y = L8(x)`,
tickmarks=[5,4],
view=[0..4,-20..0]);

Example 7.9, Page 287. Find the Laurent series for centered at .

> f := 'f': S:='S': z:='z':
f := z -> exp(-1/z^2):
S := series(f(z), z=infinity, 12):
`f(z) ` = f(z);
`f(z) ` = S;

We can get the Laurent expansion by series substitution:

> P:='P': Z:='Z':
S1 := taylor(exp(Z), Z=0, 7):
S1 := convert(S1,polynom):
`F(Z) ` = exp(Z);
P[6](Z) = S1;

Make the substitution to get the Laurent series is valid for .

> L:='L': Z:='Z':
L1 := z -> subs(Z=-1/z^2,S1):
`f(z) ` = f(z);
L[12](z) = L1(z);

Or we can get the Laurent expansion by series expansion about infinity:

> L2 := taylor(exp(-1/z^2), z=infinity, 14):
`f(z) ` = f(z);
`f(z) ` = L2;

> plot({f(x),L1(x)}, x=0.5..4.0,
title=`y = f(x), y = L12(x)`,
tickmarks=[5,4],
view=[0..4,0..1.5]);

End of Section 7.3.