Section 11

Laurent Series

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Problem 11.1

Find the set of points zz at which the Laurent series converges:

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(1)

∑n=−∞∞2−∣n∣zn\sum_{n = -\infty }^{\infty } 2^{-\left|n\right|} z^{n};

(2)

∑n=−∞∞zn3n+1\sum_{n = -\infty }^{\infty } \frac{z^{n}}{3^{n}+1};

(3)

∑n=−∞∞(z−1)nch⁡αn,α>0\sum_{n = -\infty }^{\infty } \frac{(z-1)^{n}}{\operatorname {ch} \alpha n}, \alpha > 0;

(4)

∑n=−∞∞2−n2(z+1)n\sum_{n = -\infty }^{\infty } 2^{-n^{2}}(z+1)^{n};

(5)

∑n=−∞∞(2−n3+1)−1(z−a)2n\sum_{n = -\infty }^{\infty }\left(2^{-n^{3}}+1\right)^{-1}(z-a)^{2 n};

(6)

∑n=−∞∞znn2+1\sum_{n = -\infty }^{\infty } \frac{z^{n}}{n^{2}+1};

(7)

∑n=−∞∞2−n2zn3\sum_{n = -\infty }^{\infty } 2^{-n^{2}} z^{n^{3}};

(8)

∑n=−∞∞2nzn\sum_{n = -\infty }^{\infty } 2^{n} z^{n}.

Problem 11.2

Relying on the formula for the sum of an infinitely decreasing geometric progression, and also using differentiation and integration, prove:

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(1)

1z−b=∑n=−∞−1b−n−1zn,∣z∣>∣b∣;\frac{1}{z-b} = \sum_{n = -\infty }^{-1} b^{-n-1} z^{n},\left|z\right| > \left|b\right| ;

(2)

1z2−b2=∑n=−∞−1b−2(n+1)z2n,∣z∣>∣b∣\frac{1}{z^{2}-b^{2}} = \sum_{n = -\infty }^{-1} b^{-2(n+1)} z^{2 n},\left|z\right| > \left|b\right|;

(3)

z2z2+b2=∑n=−∞0(−1)nb−2nz2n,∣z∣>∣b∣\frac{z^{2}}{z^{2}+b^{2}} = \sum_{n = -\infty }^{0}(-1)^{n} b^{-2 n} z^{2 n},\left|z\right| > \left|b\right|;

(4)

1(z−b)2=−∑n=−∞−2(n+1)b−n−2zn,∣z∣>∣b∣\frac{1}{(z-b)^{2}} = -\sum_{n = -\infty }^{-2}(n+1) b^{-n-2} z^{n}, \quad \left|z\right| > \left|b\right|;

(5)

1z−b=∑n=−∞−1(b−a)−n−1(z−a)n,a≠b,∣z−a∣>∣b−a∣\frac{1}{z-b} = \sum_{n = -\infty }^{-1}(b-a)^{-n-1}(z-a)^{n}, a \neq b,\left|z-a\right| > \left|b-a\right|;

(6)

(z−az−b)2=∑n=−∞0(1−n)(b−a)−n(z−a)n,a≠b,∣z−a∣>∣b−a∣\left(\frac{z-a}{z-b}\right)^{2} = \sum_{n = -\infty }^{0}(1-n)(b-a)^{-n}(z-a)^{n}, a \neq b,\left|z-a\right| > \left|b-a\right|.

Problem 11.3

Expand the function in a Laurent series in powers of zz in the annulus 1<∣z∣<21 < \left|z\right| < 2:

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(1)

1(z+1)(z−2)\frac{1}{(z+1)(z-2)};

(2)

z4+1(z−1)(z+2)\frac{z^{4}+1}{(z-1)(z+2)};

(3)

z(z2+1)(z+2)\frac{z}{\left(z^{2}+1\right)(z+2)};

(4)

1(z−1)2(z+2)\frac{1}{(z-1)^{2}(z+2)};

(5)

1(z2+1)(z2−4)\frac{1}{\left(z^{2}+1\right)\left(z^{2}-4\right)};

(6)

1(z2−1)2(z2+4)\frac{1}{\left(z^{2}-1\right)^{2}\left(z^{2}+4\right)}.

Problem 11.4

Expand the function in a Laurent series in powers of z−az-a in the annulus DD (the point aa and the annulus DD are indicated in parentheses):

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(1)

1z(z−3)2(a=1,D={z:1<∣z−1∣<2})\frac{1}{z(z-3)^{2}}(a = 1, \quad D = \left\{ z: 1 < \left|z-1\right| < 2\right\} );

(2)

1(z2−9)z2(a=1,D={z:1<∣z−1∣<2})\frac{1}{\left(z^{2}-9\right) z^{2}} \quad (a = 1, \quad D = \left\{ z: 1 < \left|z-1\right| < 2\right\} );

(3)

z+iz2(a=i,−i∈D)\frac{z+i}{z^{2}}(a = i,-i \in D);

(4)

z2−1z2+1(a=1,2i∈D)\frac{z^{2}-1}{z^{2}+1} \quad (a = 1, \quad 2 i \in D);

(5)

1z(z−1)(z−2)(a=0,−32∈D)\frac{1}{z(z-1)(z-2)} \quad \left(a = 0,-\frac{3}{2} \in D\right);

(6)

2zz2−2i(a=1,−1∈D)\frac{2 z}{z^{2}-2 i}(a = 1,-1 \in D);

(7)

z3(z+1)(z−2)(a=−1,D={z:0<∣z+1∣<3})\frac{z^{3}}{(z+1)(z-2)}(a = -1, \quad D = \left\{ z: 0 < \left|z+1\right| < 3\right\} );

(8)

1(z2−1)(z2+4)(a=0,D={z:∣z∣>2})\frac{1}{\left(z^{2}-1\right)\left(z^{2}+4\right)} \quad (a = 0, \quad D = \left\{ z:\left|z\right| > 2\right\} ).

Problem 11.5

Expand the given function f(z)f(z) in a Laurent series in powers of zz in the annulus to which the point z0z_{0} belongs. Indicate the boundaries of the annulus of convergence.

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(1)

f(z)=z2+1z2+32z−1,z0=1f(z) = \frac{z^{2}+1}{z^{2}+\frac{3}{2} z-1}, z_{0} = 1;

(2)

f(z)=2z2−5z2−z−2,z0=32f(z) = \frac{2 z^{2}-5}{z^{2}-z-2}, \quad z_{0} = \frac{3}{2};

(3)

f(z)=3z2+13z2−2z−1,z0=12f(z) = \frac{3 z^{2}+1}{3 z^{2}-2 z-1}, \quad z_{0} = \frac{1}{2};

(4)

f(z)=zz2+6+z−12z2−3z,z0=2f(z) = \frac{z}{z^{2}+6}+\frac{z-1}{2 z^{2}-3 z}, \quad z_{0} = 2;

(5)

f(z)=3z+12z2+z−zz2+5,z0=1f(z) = \frac{3 z+1}{2 z^{2}+z}-\frac{z}{z^{2}+5}, \quad z_{0} = 1;

(6)

f(z)=3z3+6z2−5z2+2z−3,z0=2f(z) = \frac{3 z^{3}+6 z^{2}-5}{z^{2}+2 z-3}, \quad z_{0} = 2;

(7)

f(z)=8z3+6z2−34z2+3z−1,z0=12f(z) = \frac{8 z^{3}+6 z^{2}-3}{4 z^{2}+3 z-1}, \quad z_{0} = \frac{1}{2}.

Problem 11.6

A rational function f(z)f(z) is expanded in a Laurent series in powers of zz. Expand it in a Laurent series in powers of zz, in the annulus containing the point z0z_{0}, and indicate the boundaries of the annulus of convergence, if:

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(1)

f(z)=∑n=−1−∞(n+1)3nzn+∑n=0∞(n24n+(−1)n2n)zn,z0=−3;f(z) = \sum_{n = -1}^{-\infty }(n+1) 3^{n} z^{n}+\sum_{n = 0}^{\infty }\left(\frac{n^{2}}{4^{n}}+\frac{(-1)^{n}}{2^{n}}\right) z^{n}, \quad z_{0} = -3 ;

(2)

f(z)=∑n=0∞(3nz3n+n22nz3n−1),z0=−1f(z) = \sum_{n = 0}^{\infty }\left(3^{n} z^{3 n}+\frac{n^{2}}{2^{n}} z^{3 n-1}\right), \quad z_{0} = -1;

(3)

f(z)=∑n=−1−∞(n5n+(−1)n2n)zn+∑n=0∞n23nzn,z0=13f(z) = \sum_{n = -1}^{-\infty }\left(n 5^{n}+(-1)^{n} 2^{n}\right) z^{n}+\sum_{n = 0}^{\infty } \frac{n^{2}}{3^{n}} z^{n}, \quad z_{0} = \frac{1}{3}.

Problem 11.7

Expand the function f(z)f(z) in a Laurent series in powers of z−az-a in the annulus DD to which the point z0z_{0} belongs. Indicate the boundaries of the annulus DD.

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(1)

f(z)=6z−z2z2+3z−18,a=−1,z0=3.5f(z) = \frac{6 z-z^{2}}{z^{2}+3 z-18}, a = -1, \quad z_{0} = 3.5;

(2)

f(z)=z(z+2)(z+1)2,a=−2,z0=−32f(z) = \frac{z}{(z+2)(z+1)^{2}}, \quad a = -2, \quad z_{0} = -\frac{3}{2};

(3)

f(z)=5−4z(z+1)(z2−1)2,a=1,z0=0f(z) = \frac{5-4 z}{(z+1)\left(z^{2}-1\right)^{2}}, a = 1, \quad z_{0} = 0;

(4)

f(z)=4z(z−1)(z2−1),a=−1,z0=−2f(z) = \frac{4 z}{(z-1)\left(z^{2}-1\right)}, \quad a = -1, \quad z_{0} = -2;

(5)

f(z)=9−z2z(z+9),a=−4,z0=12f(z) = \frac{9-z^{2}}{z(z+9)}, \quad a = -4, \quad z_{0} = \frac{1}{2};

(6)

f(z)=2z−1z2−z+i+1,a=3,z0=16f(z) = \frac{2 z-1}{z^{2}-z+i+1}, \quad a = 3, \quad z_{0} = \frac{1}{6};

(7)

f(z)=3z−iz2−(i+1)z−i−2,a=1,z0=52f(z) = \frac{3 z-i}{z^{2}-(i+1) z-i-2}, \quad a = 1, \quad z_{0} = \frac{5}{2}.

Problem 11.8

Expand the function f(z)f(z) in a series in powers of zz in the annulus to which the point z0z_{0} belongs. Indicate the boundaries of the annulus of convergence.

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(1)

f(z)=z2+6iz+3z2+2iz+3,z0=ief(z) = \frac{z^{2}+6 i z+3}{z^{2}+2 i z+3}, \quad z_{0} = i e;

(2)

f(z)=3iz2z2−5iz−4,z0=3if(z) = \frac{3 i z^{2}}{z^{2}-5 i z-4}, \quad z_{0} = 3 i;

(3)

f(z)=7−3z3z2−4z+1+9z2+24(3z−1)(z−2)2,z0=i+1f(z) = \frac{7-3 z}{3 z^{2}-4 z+1}+\frac{9 z^{2}+24}{(3 z-1)(z-2)^{2}}, \quad z_{0} = i+1;

(4)

f(z)=zz2−6z+8−2z2−14(z−4)(z−1)2,z0=i−1f(z) = \frac{z}{z^{2}-6 z+8}-\frac{2 z^{2}-14}{(z-4)(z-1)^{2}}, \quad z_{0} = i-1;

(5)

f(z)=(1−i)z−5iz2+(3−2i)z−6,z0=1+3if(z) = \frac{(1-i) z-5}{i z^{2}+(3-2 i) z-6}, \quad z_{0} = 1+3 i;

(6)

f(z)=(1−i)z−5iz2+(2−3i)z−6,z0=1+2if(z) = \frac{(1-i) z-5}{i z^{2}+(2-3 i) z-6}, \quad z_{0} = 1+2 i;

(7)

f(z)=3z−4(z+3)(z−2i)+z+4z2−(1+2i)z+2i,z0=1+if(z) = \frac{3 z-4}{(z+3)(z-2 i)}+\frac{z+4}{z^{2}-(1+2 i) z+2 i}, \quad z_{0} = 1+i;

(8)

f(z)=2z2+(i−2)z−2i+iz−4z2+z−6,z0=1+if(z) = \frac{2}{z^{2}+(i-2) z-2 i}+\frac{i z-4}{z^{2}+z-6}, \quad z_{0} = 1+i.

Problem 11.9

Expand the function f(z)f(z) in a Laurent series in powers of z−az-a in the annulus to which the point z0z_{0} belongs. Indicate the boundaries of the annulus of convergence.

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(1)

f(z)=−4+2i(z−1−2i)(z−5),a=1,z0=1+6if(z) = \frac{-4+2 i}{(z-1-2 i)(z-5)}, \quad a = 1, \quad z_{0} = 1+6 i;

(2)

f(z)=−4−2i(z+1+2i)(z−3),a=−1,z0=−1−5if(z) = \frac{-4-2 i}{(z+1+2 i)(z-3)}, a = -1, \quad z_{0} = -1-5 i;

(3)

f(z)=2z−3+2iz2−(1+2i)z+2i,a=2,z0=4if(z) = \frac{2 z-3+2 i}{z^{2}-(1+2 i) z+2 i}, \quad a = 2, \quad z_{0} = 4 i;

(4)

f(z)=1z2+(i−4)z+4−2i+2izz2−4,a=1,z0=2if(z) = \frac{1}{z^{2}+(i-4) z+4-2 i}+\frac{2 i z}{z^{2}-4}, \quad a = 1, \quad z_{0} = 2 i;

(5)

f(z)=5−2i−zz2+z(5+i)+5i,a=1,z0=3f(z) = \frac{5-2 i-z}{z^{2}+z(5+i)+5 i}, a = 1, \quad z_{0} = 3;

(6)

f(z)=z+2iiz2−4z+5i,a=1,z0=3f(z) = \frac{z+2 i}{i z^{2}-4 z+5 i}, \quad a = 1, \quad z_{0} = 3;

(7)

f(z)=2+15i−zz2+z(1−5i)−5i,a=−3,z0=1+if(z) = \frac{2+15 i-z}{z^{2}+z(1-5 i)-5 i}, \quad a = -3, \quad z_{0} = 1+i;

(8)

f(z)=(1+i)z+6iz2+(5+i)z+5,a=−3,z0=1+if(z) = \frac{(1+i) z+6}{i z^{2}+(5+i) z+5}, a = -3, \quad z_{0} = 1+i.

Problem 11.10

Expand the function f(z)f(z) in a Laurent series in powers of z−az-a in the annulus to which the point z0z_{0} belongs. Indicate the boundaries of the annulus of convergence.

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(1)

f(z)=z(1−i)z2−2(1+i)z+4i,a=2−3i,z0=0f(z) = \frac{z(1-i)}{z^{2}-2(1+i) z+4 i}, \quad a = 2-3 i, \quad z_{0} = 0;

(2)

f(z)=z(i−1)+i2z2−(2−i)z−i,a=2+3i,z0=0f(z) = \frac{z(i-1)+i}{2 z^{2}-(2-i) z-i}, a = 2+3 i, \quad z_{0} = 0;

(3)

f(z)=(i−3)z+4i3iz2+(1+9i)z+3,a=−2+i,z0=if(z) = \frac{(i-3) z+4 i}{3 i z^{2}+(1+9 i) z+3}, \quad a = -2+i, \quad z_{0} = i;

(4)

f(z)=(i−2)z−22iz2−(6+i)z+3,a=−i,z0=i−1f(z) = \frac{(i-2) z-2}{2 i z^{2}-(6+i) z+3}, \quad a = -i, \quad z_{0} = i-1;

(5)

f(z)=iz2+(6−i)z+9−3i+2zz2−9,a=1+i,z0=−2f(z) = \frac{i}{z^{2}+(6-i) z+9-3 i}+\frac{2 z}{z^{2}-9}, \quad a = 1+i, \quad z_{0} = -2;

(6)

f(z)=z−1−5iz2−2z+2+3z−1−3iz2−z(1+2i)−1+i,a=2i,z0=0f(z) = \frac{z-1-5 i}{z^{2}-2 z+2}+\frac{3 z-1-3 i}{z^{2}-z(1+2 i)-1+i}, a = 2 i, \quad z_{0} = 0;

(7)

f(z)=16−z2z(z+4i)2,a=−3−5i,z0=−1−if(z) = \frac{16-z^{2}}{z(z+4 i)^{2}}, \quad a = -3-5 i, \quad z_{0} = -1-i;

(8)

f(z)=2z2+iz+5z2(z−5i),a=−2−2i,z0=1+if(z) = \frac{2 z^{2}+i z+5}{z^{2}(z-5 i)}, a = -2-2 i, \quad z_{0} = 1+i;

(9)

f(z)=z+5iz2+1+6z+8iz2+3iz−2,a=1+i,z0=0f(z) = \frac{z+5 i}{z^{2}+1}+\frac{6 z+8 i}{z^{2}+3 i z-2}, a = 1+i, \quad z_{0} = 0;

(10)

f(z)=6z+8+8iz2+2z(2+i)+8i−3z−2iz2+4,a=1+i,z0=−1f(z) = \frac{6 z+8+8 i}{z^{2}+2 z(2+i)+8 i}-\frac{3 z-2 i}{z^{2}+4}, a = 1+i, \quad z_{0} = -1.

Problem 11.11

Expand the function f(z)f(z) in a Laurent series in powers of z−az-a in the annulus DD:

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(1)

f(z)=z3e1/z,a=0,D={z:0<∣z∣<∞}f(z) = z^{3} e^{1 / z}, \quad a = 0, \quad D = \left\{ z: 0 < \left|z\right| < \infty \right\};

(2)

f(z)=z2sin⁡π(z+1)z,a=0,D={z:0<∣z∣<∞}f(z) = z^{2} \sin \frac{\pi (z+1)}{z}, a = 0, D = \left\{ z: 0 < \left|z\right| < \infty \right\};

(3)

f(z)=z3cos⁡1z−2,a=0,D={z:0<∣z−2∣<∞}f(z) = z^{3} \cos \frac{1}{z-2}, a = 0, D = \left\{ z: 0 < \left|z-2\right| < \infty \right\}.

Problem 11.12

Using the expansion of the function zctg⁡zz \operatorname {ctg} z in a Taylor series (7, Example 4 (2)), expand the function ctg⁡z\operatorname {ctg} z in a Laurent series in powers of zz in the annulus π<∣z∣<2π\pi < \left|z\right| < 2 \pi.

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Problem 11.13

Using the expansion of the function zez−1\frac{z}{e^{z}-1} in a Taylor series (7, Example 4 (1)), expand the function 1ez−1\frac{1}{e^{z}-1} in a Laurent series in powers of zz in the annulus 2π<∣z∣<4π2 \pi < \left|z\right| < 4 \pi.

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Problem 11.14

Let the Laurent series ∑n=−∞∞an(z−a)n\sum_{n = -\infty }^{\infty } a_{n}(z-a)^{n} and ∑n=−∞∞bn(z−a)n\sum_{n = -\infty }^{\infty } b_{n}(z-a)^{n} converge in the annulus D={z:r<∣z−a∣<R}D = \left\{ z: r < \left|z-a\right| < R\right\}, with sums equal to f(z)f(z) and g(z)g(z), respectively. Prove that the Laurent series ∑n=−∞∞cn(z−a)n\sum_{n = -\infty }^{\infty } c_{n}(z-a)^{n}, where cn=∑k=−∞∞akbn−kc_{n} = \sum_{k = -\infty }^{\infty } a_{k} b_{n-k}, also converges in the annulus DD, and its sum equals the product f(z)g(z)f(z) g(z).

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Problem 11.15

Let the Laurent series ∑n=−∞∞cn(z−a)n\sum_{n = -\infty }^{\infty } c_{n}(z-a)^{n} converge in the closed annulus

Dˉ={z:r⩽∣z−a∣⩽R} \bar{D} = \left\{ z: r \leqslant \left|z-a\right| \leqslant R\right\}

Prove that

∣cn∣⩽M(r−n+R−n),n∈Z \left|c_{n}\right| \leqslant M\left(r^{-n}+R^{-n}\right), \quad n \in \mathbb {Z}

where MM is a constant not depending on nn.

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