Chapter 3

Laurent Series. Singular Points. Residues

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Section 11Laurent Series[15/80%]
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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Section 12Isolated Singular Points of Single-Valued Character[31/29%]
Problem 12.1

Prove that the point z=az = a is a removable singular point for the following functions:

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(1)

z2−1z−1(a=1)\frac{z^{2}-1}{z-1} \quad (a = 1);

(2)

sin⁡zz(a=0)\frac{\sin z}{z} \quad (a = 0);

(3)

ztg⁡z(a=0)\frac{z}{\operatorname {tg} z} \quad (a = 0);

(4)

1−cos⁡zz2(a=0)\frac{1-\cos z}{z^{2}}(a = 0);

(5)

ctg⁡z−1z(a=0)\operatorname {ctg} z-\frac{1}{z} \quad (a = 0);

(6)

1ez−1−1sin⁡z(a=0)\frac{1}{e^{z}-1}-\frac{1}{\sin z} \quad (a = 0);

(7)

1cos⁡2z−1(z−π2)2(a=π2)\frac{1}{\cos^{2} z}-\frac{1}{\left(z-\frac{\pi }{2}\right)^{2}} \quad \left(a = \frac{\pi }{2}\right);

(8)

z2−1z3+1(a=∞)\frac{z^{2}-1}{z^{3}+1} \quad (a = \infty ).

Problem 12.2

Prove that the point z=az = a is a pole of the following functions:

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(1)

1z(a=0)\frac{1}{z} \quad (a = 0);

(2)

1(z2+1)2(a=i)\frac{1}{\left(z^{2}+1\right)^{2}} \quad (a = i);

(3)

z2+1z+1(a=∞)\frac{z^{2}+1}{z+1} \quad (a = \infty );

(4)

z1−cos⁡z(a=0)\frac{z}{1-\cos z} \quad (a = 0);

(5)

z(ez−1)2(a=0)\frac{z}{\left(e^{z}-1\right)^{2}} \quad (a = 0);

(6)

ctg⁡πz(a=∞)\operatorname {ctg} \frac{\pi }{z} \quad (a = \infty );

(7)

z(ez+1)2(a=πi)\frac{z}{\left(e^{z}+1\right)^{2}} \quad (a = \pi i);

(8)

tg⁡πz(a=±12,±32,…)\operatorname {tg} \pi z \quad \left(a = \pm \frac{1}{2}, \pm \frac{3}{2}, \ldots \right).

Problem 12.3

Let the functions f(z)f(z) and g(z)g(z) be regular at the point z=az = a, with f(a)=g(a)=0f(a) = g(a) = 0. Prove that:

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(1)

lim⁡z→af(z)g(z)=lim⁡z→af′(z)g′(z)\lim_{z \rightarrow a} \frac{f(z)}{g(z)} = \lim_{z \rightarrow a} \frac{f^{\prime }(z)}{g^{\prime }(z)};

(2)

lim⁡z→af(z)g(z)=lim⁡z→a{g′(z)f2(z)f′(z)g2(z)}\lim_{z \rightarrow a} \frac{f(z)}{g(z)} = \lim_{z \rightarrow a}\left\{ \frac{g^{\prime }(z) f^{2}(z)}{f^{\prime }(z) g^{2}(z)}\right\}.

Problem 12.4

Let the functions f(z)f(z) and g(z)g(z) be regular at the point z=az = a, with f(a)=g(a)=0f(a) = g(a) = 0. Prove that the point z=az = a is an isolated singular point of single-valued character for the function F(z)=f(z)/g(z)F(z) = f(z) / g(z), and that it cannot be an essential singular point.

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Problem 12.5

Let z=az = a be an isolated singular point of single-valued character for the function f(z)f(z). Prove that the point z=az = a is an essential singular point of the function f(z)f(z) if and only if there exist two sequences z1′,z2′,…z_{1}^{\prime }, z_{2}^{\prime }, \ldots and z1′′,z2′′,…z_{1}^{\prime \prime }, z_{2}^{\prime \prime }, \ldots such that lim⁡n→∞zn′=lim⁡n→∞zn′′=a\lim_{n \rightarrow \infty } z_{n}^{\prime } = \lim_{n \rightarrow \infty } z_{n}^{\prime \prime } = a, while lim⁡n→∞f(zn′)=A,lim⁡n→∞f(zn′′)=B,A≠B\lim_{n \rightarrow \infty } f\left(z_{n}^{\prime }\right) = A, \lim_{n \rightarrow \infty } f\left(z_{n}^{\prime \prime }\right) = B, \quad A \neq B.

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Problem 12.6

Prove that the point z=∞z = \infty is an essential singular point for the function sin⁡z\sin z.

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Problem 12.7

Prove that the point z=az = a is an essential singular point of the functions:

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(1)

ez(a=∞);e^{z} \quad (a = \infty ) ;

(2)

e−z2(a=∞)e^{-z^{2}}(a = \infty );

(3)

sin⁡πz2(a=0)\sin \frac{\pi }{z^{2}} \quad (a = 0);

(4)

z2cos⁡πz(a=0)z^{2} \cos \frac{\pi }{z} \quad (a = 0);

(5)

etg⁡z(a=π2)e^{\operatorname {tg} z} \quad \left(a = \frac{\pi }{2}\right);

(6)

sin⁡(ez)(a=∞)\sin \left(e^{z}\right) \quad (a = \infty );

(7)

cos⁡zz+1(a=−1);\cos \frac{z}{z+1} \quad (a = -1) ;

(8)

sin⁡πz2+1(a=−i)\sin \frac{\pi }{z^{2}+1} \quad (a = -i).

Problem 12.8

Find all isolated singular points of single-valued character for the following functions and determine their type:

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(1)

zsin⁡z\frac{z}{\sin z};

(2)

1−cos⁡zsin⁡2z\frac{1-\cos z}{\sin^{2} z};

(3)

z2sin⁡zz+1z^{2} \sin \frac{z}{z+1};

(4)

1z2−1cos⁡πzz+1\frac{1}{z^{2}-1} \cos \frac{\pi z}{z+1};

(5)

ctg⁡z−1z\operatorname {ctg} z-\frac{1}{z};

(6)

z(e1/z−1)z\left(e^{1 / z}-1\right);

(7)

ectg⁡πze^{\operatorname {ctg} \frac{\pi }{z}};

(8)

sin⁡(e1/z)\sin \left(e^{1 / z}\right).

Problem 12.9

Prove that an isolated singular point of single-valued character z=az = a is a removable singular point for the function f(z)f(z), if one of the following conditions holds (arranged in order of decreasing restrictiveness):

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(1)

the function f(z)f(z) is bounded in some neighborhood of the point z=az = a;

(2)

lim⁡z→a(z−a)f(z)=0\lim_{z \rightarrow a}(z-a) f(z) = 0;

(3)

lim⁡ρ→0∫∣z−a∣=ρ∣f(z)∣∣d(z)∣=0\lim_{\rho \rightarrow 0} \int_{\left|z-a\right| = \rho }\left|f(z)\right|\left|d(z)\right| = 0.

Problem 12.10

Let z=az = a be an isolated singular point of single-valued character for a function f(z)f(z) satisfying, in some neighborhood of this point, the inequality ∣f(z)∣<M∣z−a∣−m\left|f(z)\right| < M\left|z-a\right|^{-m}, where MM and mm are some positive constants. Prove that the point z=az = a cannot be an essential singular point of the function f(z)f(z).

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Problem 12.11

Let the function f(z)f(z) have a pole of order mm at the finite point z=az = a. Determine the order of the pole at the point z=az = a of the function f(n)(z)f^{(n)}(z).

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Problem 12.12

Let Pm(z)P_{m}(z) and Qn(z)Q_{n}(z) be polynomials of degree mm and nn, respectively. Find the order of the pole at infinity for the following functions:

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(1)

Pm(z)+Qn(z),m≠nP_{m}(z)+Q_{n}(z), m \neq n;

(2)

Pm(z)⋅Qn(z)P_{m}(z) \cdot Q_{n}(z);

(3)

Pm(z)Qn(z),m>n\frac{P_{m}(z)}{Q_{n}(z)}, m > n.

Problem 12.13

Let the functions f(z)f(z) and g(z)g(z) have, at the point z=∞z = \infty, poles of orders mm and nn, respectively. Prove that the function F(z)=f(g(z))F(z) = f(g(z)) has, at the point z=∞z = \infty, a pole of order mnm n.

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Problem 12.14

Let the function g(z)g(z) be regular at the point z=az = a with g(a)=bg(a) = b, and let the function f(ζ)f(\zeta ) have a pole of order mm at the point ζ=b\zeta = b. Prove that the function F(z)=f(g(z))F(z) = f(g(z)) has, at the point z=az = a, a pole of order mnm n, where nn is the multiplicity of the zero of the function g(z)−bg(z)-b at the point z=az = a.

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Problem 12.15

Find all singular points of the function f(z)f(z) and determine their type, if:

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(1)

f(z)=1−ch⁡z2ez−e3zf(z) = \frac{1-\operatorname {ch} \frac{z}{2}}{e^{z}-e^{3 z}};

(2)

f(z)=z2−4z−5(z+3)(1+cos⁡πz)f(z) = \frac{z^{2}-4 z-5}{(z+3)(1+\cos \pi z)};

(3)

f(z)=e2z−ez(z2−1)sin⁡πzf(z) = \frac{e^{2 z}-e^{z}}{\left(z^{2}-1\right) \sin \pi z};

(4)

f(z)=14sh⁡(z−1)+1z2−6z+5f(z) = \frac{1}{4 \operatorname {sh}(z-1)}+\frac{1}{z^{2}-6 z+5};

(5)

f(z)=6sh⁡z−z(6+z2)(ez−1)5f(z) = \frac{6 \operatorname {sh} z-z\left(6+z^{2}\right)}{\left(e^{z}-1\right)^{5}};

(6)

f(z)=e3z−cos⁡2zz2−3cth⁡zf(z) = \frac{e^{3 z}-\cos^{2} z}{z^{2}}-3 \operatorname {cth} z;

(7)

f(z)=etg⁡πzsin⁡2πz3(2z+1)(ez+1)f(z) = \frac{e^{\operatorname {tg} \pi z} \sin \frac{2 \pi z}{3}}{(2 z+1)\left(e^{z}+1\right)};

(8)

f(z)=tg⁡zetg⁡ztg⁡4zf(z) = \frac{\operatorname {tg} z e^{\operatorname {tg} z}}{\operatorname {tg} 4 z};

(9)

f(z)=z3−sh3z(ez−1)5f(z) = \frac{z^{3}-\mathrm{sh}^{3} z}{\left(e^{z}-1\right)^{5}};

(10)

f(z)=ectg⁡πzcos⁡πz4(z−1)2(ch⁡z+1)f(z) = \frac{e^{\operatorname {ctg} \pi z} \cos \frac{\pi z}{4}}{(z-1)^{2}(\operatorname {ch} z+1)};

(11)

f(z)=4z2+4πz−8π2(e2iz−1)2−3πz+2πf(z) = \frac{4 z^{2}+4 \pi z-8 \pi^{2}}{\left(e^{2 i z}-1\right)^{2}}-\frac{3 \pi }{z+2 \pi };

(12)

f(z)=z3+2πz2+π2zsin⁡3z−πz+πf(z) = \frac{z^{3}+2 \pi z^{2}+\pi^{2} z}{\sin^{3} z}-\frac{\pi }{z+\pi }.

Problem 12.16

Let a sequence of functions {fn(z)}\left\{ f_{n}(z)\right\}, regular in the annulus 0<∣z−a∣<r0 < \left|z-a\right| < r, converge uniformly on every closed part of this annulus. Prove that if each function fn(z)f_{n}(z) has a pole of order mm at the point z=az = a, then the limit function f(z)f(z) is regular in the annulus 0<∣z−a∣<r0 < \left|z-a\right| < r and has, at the point z=az = a, either a pole of order not exceeding mm, or a removable singular point.

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Problem 12.17

Find all singular points of the function f(z)f(z) and determine their type, if:

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(1)

f(z)=sin⁡πzz2+z−2f(z) = \frac{\sin \frac{\pi }{z}}{z^{2}+z-2};

(2)

f(z)=cos⁡πzsin⁡z+1f(z) = \frac{\cos \frac{\pi }{z}}{\sin z+1};

(3)

f(z)=1−sin⁡iπzeπz−1f(z) = \frac{1-\sin \frac{i \pi }{z}}{e^{\pi z}-1};

(4)

f(z)=z2+4z−121+sin⁡3πzf(z) = \frac{z^{2}+4 z-12}{1+\sin \frac{3 \pi }{z}};

(5)

f(z)=8z2−6z+11−cos⁡πzf(z) = \frac{8 z^{2}-6 z+1}{1-\cos \frac{\pi }{z}};

(6)

f(z)=cos⁡πzz+11+cos⁡πzf(z) = \frac{\cos \frac{\pi z}{z+1}}{1+\cos \pi z};

(7)

f(z)=(z−2)(3z2−4z−4)1−sin⁡πzf(z) = \frac{(z-2)\left(3 z^{2}-4 z-4\right)}{1-\sin \frac{\pi }{z}};

(8)

f(z)=sin⁡πzeπ/(z+i)−if(z) = \frac{\sin \pi z}{e^{\pi /(z+i)}-i};

(9)

f(z)=e1/(z−2i)1−cos⁡iπzf(z) = \frac{e^{1 /(z-2 i)}}{1-\cos i \pi z};

(10)

f(z)=ecos⁡(πi/2z)−1i+sh⁡3πz2f(z) = \frac{e^{\cos (\pi i / 2 z)}-1}{i+\operatorname {sh} \frac{3 \pi z}{2}}.

Problem 12.18

Let z=az = a be a pole of the function f(z)f(z). Prove that for the function ef(z)e^{f(z)} the point z=az = a is an essential singular point.

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Problem 12.19

Let f(z)f(z) be a function, regular on the set C‾\overline{\mathbb {C}}, except for a single simple pole z=az = a. Prove that if a=∞a = \infty, then f(z)f(z) is a linear function, i.e., f(z)=Az+Bf(z) = A z+B, where A≠0A \neq 0, and if a≠∞a \neq \infty, then f(z)f(z) is a fractional-linear function, i.e., f(z)=az+bcz+df(z) = \frac{a z+b}{c z+d}, where ad−bc≠0a d-b c \neq 0, c≠0c \neq 0.

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Problem 12.20

Find all singular points of the function f(z)f(z) and determine their type, if:

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(1)

f(z)=z2+ln⁡22sin⁡z−54ch⁡1zf(z) = \frac{z^{2}+\ln^{2} 2}{\sin z-\frac{5}{4}} \operatorname {ch} \frac{1}{z};

(2)

f(z)=4z2+π21+e2z(ch⁡πz−1)f(z) = \frac{4 z^{2}+\pi^{2}}{1+e^{2 z}}\left(\operatorname {ch} \frac{\pi }{z}-1\right);

(3)

f(z)=z2+ln⁡22cos⁡z+3i4sh⁡1zf(z) = \frac{z^{2}+\ln^{2} 2}{\cos z+\frac{3 i}{4}} \operatorname {sh} \frac{1}{z};

(4)

f(z)=4z2−12iz−5sh⁡πz−ie1/(sh⁡π/z)f(z) = \frac{4 z^{2}-12 i z-5}{\operatorname {sh} \pi z-i} e^{1 /(\operatorname {sh} \pi / z)};

(5)

f(z)=sh⁡πz−cos⁡iπz(i−eπ/z)2f(z) = \frac{\operatorname {sh} \pi z-\cos \frac{i \pi }{z}}{\left(i-e^{\pi / z}\right)^{2}};

(6)

f(z)=(2z−π)e1/cos⁡zzcos⁡2zcos⁡zf(z) = \frac{(2 z-\pi ) e^{1 / \cos z}}{z \cos 2 z \cos z};

(7)

f(z)=cos⁡πi2(z−1)eπz+eπf(z) = \frac{\cos \frac{\pi i}{2(z-1)}}{e^{\pi z}+e^{\pi }};

(8)

f(z)=4z2−12z+5sin⁡πz−1e1/(sin⁡π/z)f(z) = \frac{4 z^{2}-12 z+5}{\sin \pi z-1} e^{1 /(\sin \pi / z)};

(9)

f(z)=e1/sin⁡z(z2+4π2)z2sh⁡zf(z) = \frac{e^{1 / \sin z}\left(z^{2}+4 \pi^{2}\right)}{z^{2} \operatorname {sh} z};

(10)

f(z)=z3sh⁡14ze1/z−e2/zf(z) = \frac{z^{3} \operatorname {sh} \frac{1}{4 z}}{e^{1 / z}-e^{2 / z}}.

Problem 12.21

Let f(z)f(z) be an entire function taking each finite value only once. Prove that f(z)f(z) is a linear function, i.e., f(z)=az+bf(z) = a z+b, a≠0a \neq 0.

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Problem 12.22

Let z=az = a be an essential singular point of the function f(z)f(z). Can one assert that for the function 1f(z)\frac{1}{f(z)} the point z=az = a is also essential?

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Problem 12.23

Let the point z=az = a be an essential singular point of the function f(z)f(z). Denote M(ρ)=max⁡∣z−a∣=ρ∣f(z)∣M(\rho ) = \max_{\left|z-a\right| = \rho }\left|f(z)\right|. Prove that for any k>0k > 0

lim⁡ρ→0ρkM(ρ)=+∞ \lim _{\rho \rightarrow 0} \rho ^{k} M(\rho ) = +\infty
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Problem 12.24

Let the function g(z)g(z) be regular at the point z=az = a with g(a)=bg(a) = b, and let ζ=b\zeta = b be an essential singular point of the function f(ζ)f(\zeta ). Prove that z=az = a is an essential singular point of the function F(z)=f(g(z))F(z) = f(g(z)).

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Problem 12.25

Let the functions a1(z),…,an(z)a_{1}(z), \ldots , a_{n}(z) be regular at the point aa or have a pole there, and let z=az = a be an essential singular point of the function f(z)f(z). Prove that z=az = a is an essential singular point of the function

F(z)=fn(z)+a1(z)fn−1(z)+…+an(z) F(z) = f^{n}(z)+a_{1}(z) f^{n-1}(z)+\ldots +a_{n}(z)
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Problem 12.26

Find all singular points of the function f(z)f(z) and determine their type, if:

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(1)

f(z)=ctg⁡ez−1ez+1f(z) = \operatorname {ctg} \frac{e^{z}-1}{e^{z}+1};

(2)

f(z)=1sh⁡(πi1+ez1−ez)f(z) = \frac{1}{\operatorname {sh}\left(\pi i \frac{1+e^{z}}{1-e^{z}}\right)};

(3)

f(z)=z2sin⁡2zsin⁡(z−(π/2))−1(cos⁡z−1)2esin⁡z/zf(z) = \frac{z^{2} \sin^{2} z \sin (z-(\pi / 2))^{-1}}{(\cos z-1)^{2}} e^{\sin z / z};

(4)

f(z)=e1/ch⁡2z(πz+π+1)1+cos⁡11+zf(z) = \frac{e^{1 / \operatorname {ch} 2 z}(\pi z+\pi +1)}{1+\cos \frac{1}{1+z}};

(5)

f(z)=ecos⁡z/2/(1+cos⁡z)(z+π)3(1+cos⁡2z)f(z) = \frac{e^{\cos z / 2 /(1+\cos z)}}{(z+\pi )^{3}\left(1+\cos^{2} z\right)};

(6)

f(z)=z2+π2sh⁡2z(eπ/z−e2)tg⁡2zf(z) = \frac{z^{2}+\pi^{2}}{\operatorname {sh} 2 z}\left(e^{\pi / z}-e^{2}\right) \operatorname {tg}^{2} z;

(7)

f(z)=esin⁡z/(1−cos⁡z)(2π−z)2(1+sin⁡2z)f(z) = \frac{e^{\sin z /(1-\cos z)}}{(2 \pi -z)^{2}\left(1+\sin^{2} z\right)};

(8)

f(z)=4z2+π2ch⁡3z(ez/(z−2)−1)ctg⁡2zf(z) = \frac{4 z^{2}+\pi^{2}}{\operatorname {ch} 3 z}\left(e^{z /(z-2)}-1\right) \operatorname {ctg}^{2} z;

(9)

f(z)=zsin⁡3zcos⁡11−z(cos⁡z−1)2esin⁡2z/z2f(z) = \frac{z \sin^{3} z \cos \frac{1}{1-z}}{(\cos z-1)^{2}} e^{\sin^{2} z / z^{2}};

(10)

f(z)=(z−π2)3cos⁡zcos⁡1z−π(sin⁡z−1)2ecos⁡z/(z−(π/2))f(z) = \frac{\left(z-\frac{\pi }{2}\right)^{3} \cos z \cos \frac{1}{z-\pi }}{(\sin z-1)^{2}} e^{\cos z /(z-(\pi / 2))}.

Problem 12.27

Let Pm(z)P_{m}(z) and Qn(z)Q_{n}(z) be polynomials of degree mm and nn, respectively, having no common zeros. Prove that the zeros of the polynomial Qn(z)Q_{n}(z), and only these points, are poles of the function f(z)=Pm(z)Qn(z)f(z) = \frac{P_{m}(z)}{Q_{n}(z)}, and the function f(z)f(z) has no other singular points in the finite plane. Show that the point z=∞z = \infty is a pole of order m−nm-n of the function f(z)f(z) if m>nm > n, and a point of regularity if m⩽nm \leqslant n.

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Problem 12.28

Let a meromorphic function f(z)f(z) have, in C‾\overline{\mathbb {C}}, only a finite number of poles a1,a2,…,ama_{1}, a_{2}, \ldots , a_{m} (the point z=∞z = \infty may also be a pole). Prove that f(z)f(z) is a rational function (a ratio of two polynomials) and is represented in the form

f(z)=A+f0(z)+∑k=1mfk(z), f(z) = A+f_{0}(z)+\sum _{k = 1}^{m} f_{k}(z),

where fk(z)f_{k}(z) and f0(z)f_{0}(z) are, respectively, the principal parts of the Laurent series of the function f(z)f(z) in the neighborhoods of the points aka_{k} and z=∞,A=lim⁡z→∞(f(z)−f0(z))z = \infty , A = \lim_{z \rightarrow \infty }\left(f(z)-f_{0}(z)\right).

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Problem 12.29

Let the function f(z)f(z) be regular in the domain D={z:0<∣z∣<ρ}D = \left\{ z: 0 < \left|z\right| < \rho \right\} and f(z)≢0f(z) \not\equiv 0. What conditions must the function f(z)f(z) satisfy at the point z=0z = 0 so that one can specify a function g(z)g(z), regular in the disk Bρ(0)={z:∣z∣<ρ}B_{\rho }(0) = \left\{ z:\left|z\right| < \rho \right\}, with g(z)≢0g(z) \not\equiv 0, such that lim⁡z→0[f(z)g(z)]=0\lim_{z \rightarrow 0}[f(z) g(z)] = 0.

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Problem 12.30

Let the function f(z)f(z) be regular in a punctured neighborhood of the point z=0z = 0, i.e., in the domain D={z:0<∣z∣<ρ}D = \left\{ z: 0 < \left|z\right| < \rho \right\}, and let the inequality

∣f(z)∣⩽A∣z∣, where A>0. \left|f(z)\right| \leqslant \frac{A}{\sqrt{\left|z\right|}}, \quad \text{ where } \quad A > 0.

hold for all z∈Dz \in D, where A>0A > 0. Prove that z=0z = 0 is a removable singular point of the function f(z)f(z).

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Problem 12.31

Let an entire function f(z)f(z) satisfy, in some domain ∣z∣>R\left|z\right| > R, the condition ∣f(z)∣⩾A∣z∣m\left|f(z)\right| \geqslant A\left|z\right|^{m}, where A>0A > 0 and m⩾0m \geqslant 0 is an integer. Prove that f(z)f(z) is a polynomial.

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Section 13Computing Residues[18/33%]
Problem 13.1

Compute:

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(1)

resz=∞sin⁡zz2\mathop{\mathrm{res}}_{z = \infty } \frac{\sin z}{z^{2}};

(2)

resz=∞e1/z\mathop{\mathrm{res}}_{z = \infty } e^{1 / z};

(3)

resz=1ez(z−1)2\mathop{\mathrm{res}}_{z = 1} \frac{e^{z}}{(z-1)^{2}};

(4)

resz=∞z2sin⁡πz\mathop{\mathrm{res}}_{z = \infty } z^{2} \sin \frac{\pi }{z};

(5)

resz=π/4cos⁡zz−π4\mathop{\mathrm{res}}_{z = \pi / 4} \frac{\cos z}{z-\frac{\pi }{4}};

(6)

resz=1ze1/(z−1)\mathop{\mathrm{res}}_{z = 1} z e^{1 /(z-1)}.

Problem 13.2

Find the residues of the following functions at all their finite singular points:

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(1)

1z+z3\frac{1}{z+z^{3}};

(2)

z21+z4\frac{z^{2}}{1+z^{4}};

(3)

z2(1+z)3\frac{z^{2}}{(1+z)^{3}};

(4)

1(z2+1)3\frac{1}{\left(z^{2}+1\right)^{3}};

(5)

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

(6)

z2n(z−1)n,n∈N\frac{z^{2 n}}{(z-1)^{n}}, n \in \mathbb {N};

(7)

1sin⁡πz\frac{1}{\sin \pi z};

(8)

ctg⁡πz\operatorname {ctg} \pi z;

(9)

th⁡z\operatorname {th} z;

(10)

cth⁡2πz\operatorname {cth}^{2} \pi z;

(11)

cos⁡z(z−1)2\frac{\cos z}{(z-1)^{2}};

(12)

1ez+1\frac{1}{e^{z}+1};

(13)

sin⁡πz(z−1)3\frac{\sin \pi z}{(z-1)^{3}};

(14)

1sin⁡z2\frac{1}{\sin z^{2}}.

Problem 13.3

Find resz=af(z)\mathop{\mathrm{res}}_{z = a} f(z), if:

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(1)

f(z)=ez+1(z+1)2,a=∞;f(z) = \frac{e^{z+1}}{(z+1)^{2}}, \quad a = \infty ;

(2)

f(z)=1(z2+4)2z3,a=0f(z) = \frac{1}{\left(z^{2}+4\right)^{2} z^{3}}, \quad a = 0;

(3)

f(z)=sin⁡(1z+1z2),a=0f(z) = \sin \left(\frac{1}{z}+\frac{1}{z^{2}}\right), a = 0;

(4)

f(z)=z2cos⁡(1z−1z2),a=0f(z) = z^{2} \cos \left(\frac{1}{z}-\frac{1}{z^{2}}\right), a = 0;

(5)

f(z)=z2sin⁡zz−1,a=1f(z) = z^{2} \sin \frac{z}{z-1}, \quad a = 1;

(6)

f(z)=3z2+4z+52z2−3z+4,a=∞f(z) = \frac{3 z^{2}+4 z+5}{2 z^{2}-3 z+4}, \quad a = \infty.

Problem 13.4

Find the residues of the following functions at infinity:

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(1)

z4+1z6−1\frac{z^{4}+1}{z^{6}-1};

(2)

cos⁡πz+22z\cos \pi \frac{z+2}{2 z};

(3)

sin⁡1zz−1\frac{\sin \frac{1}{z}}{z-1};

(4)

cos⁡2πzz+1\frac{\cos^{2} \frac{\pi }{z}}{z+1};

(5)

(z10+1)cos⁡1z(z5+2)(z6−1)\frac{\left(z^{10}+1\right) \cos \frac{1}{z}}{\left(z^{5}+2\right)\left(z^{6}-1\right)};

(6)

zcos⁡2πzz \cos^{2} \frac{\pi }{z}.

Problem 13.5

Find the residues of the following functions at all their singular points and at infinity:

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(1)

1z6(z−2)\frac{1}{z^{6}(z-2)};

(2)

1+z8z6(z+2)\frac{1+z^{8}}{z^{6}(z+2)};

(3)

1+z10z6(z2+4)\frac{1+z^{10}}{z^{6}\left(z^{2}+4\right)};

(4)

1+z2nzn(z−a),a≠0,n∈N\frac{1+z^{2 n}}{z^{n}(z-a)}, a \neq 0, n \in \mathbb {N};

(5)

sin⁡zsin⁡1z\sin z \sin \frac{1}{z};

(6)

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

(7)

1+z8z4(z4+1)cos⁡zch⁡z\frac{1+z^{8}}{z^{4}\left(z^{4}+1\right)} \cos z \operatorname {ch} z;

(8)

sin⁡z(z2+1)2\frac{\sin z}{\left(z^{2}+1\right)^{2}}.

Problem 13.6

Compute:

?
(1)

resz=0zn−1sin⁡nz,n∈N\mathop{\mathrm{res}}_{z = 0} \frac{z^{n-1}}{\sin^{n} z}, n \in \mathbb {N};

(2)

resz=0sin⁡3z−3sin⁡zsin⁡z(sin⁡z−z)\mathop{\mathrm{res}}_{z = 0} \frac{\sin 3 z-3 \sin z}{\sin z(\sin z-z)};

(3)

resz=0tg⁡z−z(1−cos⁡z)2\mathop{\mathrm{res}}_{z = 0} \frac{\operatorname {tg} z-z}{(1-\cos z)^{2}};

(4)

resz=0zn−2sh⁡nz,n∈N,n≠1\mathop{\mathrm{res}}_{z = 0} \frac{z^{n-2}}{\operatorname {sh}^{n} z}, n \in \mathbb {N}, n \neq 1;

(5)

resz=0zn−3ctg⁡nz,n∈N,n≠1;\mathop{\mathrm{res}}_{z = 0} z^{n-3} \operatorname {ctg}^{n} z, n \in \mathbb {N}, n \neq 1 ;

(6)

resz=0zch⁡z−1−z2/2\mathop{\mathrm{res}}_{z = 0} \frac{z}{\operatorname {ch} z-1-z^{2} / 2}.

Problem 13.7

Let the function g(z)g(z) be regular at the point z=∞z = \infty, and let the function h(z)h(z) have a zero of multiplicity 1 at the point z=∞z = \infty. Prove that

resz=∞[g(z)h(z)]=−g(∞)lim⁡z→∞(zh(z)) \mathop{\mathrm{res}}_{z = \infty }[g(z) h(z)] = -g(\infty ) \lim _{z \rightarrow \infty }(z h(z))
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Problem 13.8

Prove that for every even function f(z)f(z)

resz=0f(z)=resz=∞f(z)=0 \mathop{\mathrm{res}}_{z = 0} f(z) = \mathop{\mathrm{res}}_{z = \infty } f(z) = 0

(assuming that the residues written make sense).

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Problem 13.9

Prove that for an even function f(z)f(z) the equality holds

resz=z0f(z)=−resz=−z0f(z) \mathop{\mathrm{res}}_{z = z_{0}} f(z) = -\mathop{\mathrm{res}}_{z = -z_{0}} f(z)

and for an odd function f(z)f(z), the equality

resz=z0f(z)=resz=−z0f(z) \mathop{\mathrm{res}}_{z = z_{0}} f(z) = \mathop{\mathrm{res}}_{z = -z_{0}} f(z)

(assuming that the residues written make sense).

?
Problem 13.10

Let f(z)=g(az)f(z) = g(a z), where a≠0a \neq 0. Prove that

resz=z0af(z)=1aresz=z0g(z) \mathop{\mathrm{res}}_{z = z_{0} a} f(z) = \frac{1}{a} \mathop{\mathrm{res}}_{z = z_{0}} g(z)
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Problem 13.11

Let f(z)=zmg(zn)f(z) = z^{m} g\left(z^{n}\right), where mm and nn are integers satisfying the conditions m⩾0,m<nm \geqslant 0, m < n. Prove that

resz=z0e2kπi/nf(z)=e2kπi(m+1)/nresz=z0f(z). \mathop{\mathrm{res}}_{z = z_{0} e^{2 k \pi i / n}} f(z) = e^{2 k \pi i(m+1) / n} \mathop{\mathrm{res}}_{z = z_{0}} f(z).
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Problem 13.12

Let the functions φ(z)\varphi (z) and ψ(z)\psi (z) be regular at a finite point z0z_{0} and have a zero of order mm there. Prove the formulas:

?
(1)

resz=z0{φ(z)ψ(z)⋅1z−z0}=φ(m)(z0)ψ(m)(z0);\mathop{\mathrm{res}}_{z = z_{0}}\left\{ \frac{\varphi (z)}{\psi (z)} \cdot \frac{1}{z-z_{0}}\right\} = \frac{\varphi^{(m)}\left(z_{0}\right)}{\psi^{(m)}\left(z_{0}\right)} ;

(2)

resz=z0{φ(z)ψ(z)⋅1(z−z0)2}=1m+1φ(m)(z0)ψ(m)(z0)[φ(m+1)(z0)φ(m)(z0)−ψ(m+1)(z0)ψ(m)(z0)]\mathop{\mathrm{res}}_{z = z_{0}}\left\{ \frac{\varphi (z)}{\psi (z)} \cdot \frac{1}{\left(z-z_{0}\right)^{2}}\right\} = \frac{1}{m+1} \frac{\varphi^{(m)}\left(z_{0}\right)}{\psi^{(m)}\left(z_{0}\right)}\left[\frac{\varphi^{(m+1)}\left(z_{0}\right)}{\varphi^{(m)}\left(z_{0}\right)}-\frac{\psi^{(m+1)}\left(z_{0}\right)}{\psi^{(m)}\left(z_{0}\right)}\right].

Problem 13.13

Find resz=z0{f(z)φ(z)}\mathop{\mathrm{res}}_{z = z_{0}}\left\{ f(z) \varphi (z)\right\}, if the function φ(z)\varphi (z) is regular at the point z0z_{0}, and the function f(z)f(z) has a simple pole at the point z0z_{0} with residue AA.

?
Problem 13.14

Find resz=z0f′(z)f(z)\mathop{\mathrm{res}}_{z = z_{0}} \frac{f^{\prime }(z)}{f(z)}, if the function f(z)f(z):

?
(1)

is regular at the point z0z_{0} and has a zero of multiplicity mm there;

(2)

has a pole of order mm at the point z0z_{0}.

Problem 13.15

Find resz=z0{φ(z)f′(z)f(z)}\mathop{\mathrm{res}}_{z = z_{0}}\left\{ \varphi (z) \frac{f^{\prime }(z)}{f(z)}\right\}, if the function φ(z)\varphi (z) is regular at the point z=z0z = z_{0}, and the function f(z)f(z):

?
(1)

is regular at the point z0z_{0} and has a zero of multiplicity mm there;

(2)

has a pole of order mm at the point z0z_{0}.

Problem 13.16

Find resz=z0f(φ(z))\mathop{\mathrm{res}}_{z = z_{0}} f(\varphi (z)), if the function φ(z)\varphi (z) is regular at the point z0z_{0} with φ′(z0)≠0\varphi^{\prime }\left(z_{0}\right) \neq 0, and the function f(z)f(z) has a simple pole at the point φ(z0)\varphi \left(z_{0}\right) with residue AA.

?
Problem 13.17

Let the function f(z)f(z) be regular for R<∣z∣<∞R < \left|z\right| < \infty, with resz=∞f(z)=0\mathop{\mathrm{res}}_{z = \infty } f(z) = 0. Prove that the function f(z)f(z) has an antiderivative, regular for R<∣z∣<∞R < \left|z\right| < \infty.

?
Problem 13.18

Determine for which values of the parameters a,ba, b the following functions are single-valued in the entire complex plane:

?
(1)

∫1za+e1/ζζdζ\int_{1}^{z} \frac{a+e^{1 / \zeta }}{\zeta } d \zeta;

(2)

∫1zasin⁡ζ+bζcos⁡ζζ2dζ\int_{1}^{z} \frac{a \sin \zeta +b \zeta \cos \zeta }{\zeta^{2}} d \zeta;

(3)

∫1zsin⁡(ζ+aζ3)ζ4dζ\int_{1}^{z} \frac{\sin \left(\zeta +a \zeta^{3}\right)}{\zeta^{4}} d \zeta.

Section 14Computing Integrals Over a Closed Contour[10/50%]
Problem 14.1

Compute the integrals of rational functions:

?
(1)

∮∣z∣=1dzz\oint_{\left|z\right| = 1} \frac{d z}{z};

(2)

∮∣z∣=2dz(z−1)2\oint_{\left|z\right| = 2} \frac{d z}{(z-1)^{2}};

(3)

∮∣z∣=2dzz2+1\oint_{\left|z\right| = 2} \frac{d z}{z^{2}+1};

(4)

∮∣z∣=3dzz2+z+1\oint_{\left|z\right| = 3} \frac{d z}{z^{2}+z+1};

(5)

∮∣z∣=3z4z2+z−2dz\oint_{\left|z\right| = 3} \frac{z^{4}}{z^{2}+z-2} d z;

(6)

∮∣z+i∣=2dzz2(z2−7z+12)\oint_{\left|z+i\right| = 2} \frac{d z}{z^{2}\left(z^{2}-7 z+12\right)}.

Problem 14.2

Compute the integrals:

?
(1)

∮∣z∣=3zcos⁡πzz+2idz\oint_{\left|z\right| = 3} \frac{z \cos \frac{\pi }{z}}{z+2 i} d z;

(2)

∮∣z∣=4e1/(z−1)z−2dz\oint_{\left|z\right| = 4} \frac{e^{1 /(z-1)}}{z-2} d z;

(3)

∮∣z−12∣=1z3e1/z1−z2dz\oint_{\left|z-\frac{1}{2}\right| = 1} \frac{z^{3} e^{1 / z}}{1-z^{2}} d z;

(4)

∮∣z∣=2z2sin⁡11−zdz\oint_{\left|z\right| = 2} z^{2} \sin \frac{1}{1-z} d z;

(5)

∮∣z∣=114i+3zcos⁡5i6z+3idz\oint_{\left|z\right| = 1} \frac{1}{4 i+3 z} \cos \frac{5 i}{6 z+3 i} d z;

(6)

∮∣z∣=213+ize4i/(z−i)dz\oint_{\left|z\right| = 2} \frac{1}{3+i z} e^{4 i /(z-i)} d z;

(7)

∮∣z∣=3/2ztg⁡z(z2−1)2dz\oint_{\left|z\right| = 3 / 2} \frac{z \operatorname {tg} z}{\left(z^{2}-1\right)^{2}} d z;

(8)

∮∣z∣=5/2z2z−3sin⁡zz−2dz\oint_{\left|z\right| = 5 / 2} \frac{z^{2}}{z-3} \sin \frac{z}{z-2} d z;

(9)

∮∣z−i∣=3zz2+9ch⁡zz−2dz\oint_{\left|z-i\right| = 3} \frac{z}{z^{2}+9} \operatorname {ch} \frac{z}{z-2} d z;

(10)

∮∣z∣=6z2eiz+idz\oint_{\left|z\right| = 6} \frac{z^{2}}{e^{i z}+i} d z;

(11)

∮∣z∣=3dz(z−1)2(1−e2z)\oint_{\left|z\right| = 3} \frac{d z}{(z-1)^{2}\left(1-e^{2 z}\right)};

(12)

∮∣z∣=2z2z−1e1/(z−1)dz\oint_{\left|z\right| = 2} \frac{z^{2}}{z-1} e^{1 /(z-1)} d z;

(13)

∮∣z+13∣=163z+11−sin⁡π2zdz\oint_{\left|z+\frac{1}{3}\right| = \frac{1}{6}} \frac{3 z+1}{1-\sin \frac{\pi }{2 z}} d z;

(14)

∮∣z∣=4zdze1/z+e1/2z\oint_{\left|z\right| = 4} \frac{z d z}{e^{1 / z}+e^{1 / 2 z}};

(15)

∮∣z∣=2ez5z(1+z5)dz\oint_{\left|z\right| = 2} \frac{e^{z^{5}}}{z\left(1+z^{5}\right)} d z;

(16)

∮∣z−π2(1−i)∣=πzdzcos⁡z−ch⁡z\oint_{\left|z-\frac{\pi }{2}(1-i)\right| = \pi } \frac{z d z}{\cos z-\operatorname {ch} z};

(17)

∮∣z∣=1(z−i)sin⁡1iz(z−3i)2dz\oint_{\left|z\right| = 1} \frac{(z-i) \sin \frac{1}{i z}}{(z-3 i)^{2}} d z;

(18)

∮∣z∣=1(z+1)cos⁡iz(2i−z)2dz\oint_{\left|z\right| = 1} \frac{(z+1) \cos \frac{i}{z}}{(2 i-z)^{2}} d z;

(19)

∮∣z∣=1/2z+2π(2sin⁡13z−1)dz\oint_{\left|z\right| = 1 / 2} \frac{z+\frac{2}{\pi }}{\left(2 \sin \frac{1}{3 z}-1\right)} d z;

(20)

∮∣z+i∣=3/22(z+3i)ch⁡12zz4+10z2+9dz\oint_{\left|z+i\right| = 3 / 2} \frac{2(z+3 i) \operatorname {ch} \frac{1}{2 z}}{z^{4}+10 z^{2}+9} d z;

(21)

∮∣z∣=1z−1ln⁡2e1/z−2dz\oint_{\left|z\right| = 1} \frac{z-\frac{1}{\ln 2}}{e^{1 / z}-2} d z;

(22)

∮∣z−π2∣=32z−iln⁡2tg⁡z−i3dz\oint_{\left|z-\frac{\pi }{2}\right| = 3} \frac{2 z-i \ln 2}{\operatorname {tg} z-\frac{i}{3}} d z;

(23)

∮∣z+1−i∣=2z+i(z−i)sh⁡12zdz\oint_{\left|z+1-i\right| = 2} \frac{z+i}{(z-i) \operatorname {sh} \frac{1}{2 z}} d z;

(24)

∮∣z−1∣=1zdz(π−3z)(1+cos⁡3z)\oint_{\left|z-1\right| = 1} \frac{z d z}{(\pi -3 z)(1+\cos 3 z)}.

Problem 14.3

Making an appropriate change of variable, reduce the given integral to an integral over a closed contour in C\mathbb {C} and compute it:

?
(1)

∫02πdφa+cos⁡φ(a>1)\int_{0}^{2 \pi } \frac{d \varphi }{a+\cos \varphi }(a > 1);

(2)

∫02πdφ(a+bcos⁡φ)2(a>b>0)\int_{0}^{2 \pi } \frac{d \varphi }{(a+b \cos \varphi )^{2}} \quad (a > b > 0);

(3)

∫02πdφ(a+bcos⁡2φ)2\int_{0}^{2 \pi } \frac{d \varphi }{\left(a+b \cos^{2} \varphi \right)^{2}};

(4)

∫02πecos⁡φcos⁡(nφ−sin⁡φ)dφ,n∈Z\int_{0}^{2 \pi } e^{\cos \varphi } \cos (n \varphi -\sin \varphi ) d \varphi , \quad n \in \mathbb {Z}.

Remark.
?

In Problems 4--6 it is assumed that, while moving along the boundary Γ\Gamma of the domain DD in the direction of integration, the domain DD itself remains on the left.

Problem 14.4

Prove that the integral

∫Γeiz(z−i)2dz \int _{\Gamma } \frac{e^{i z}}{(z-i)^{2}} d z

taken over the boundary Γ\Gamma of the half-plane {z:Im(z)>0}\left\{ z: \mathrm{Im}\left(z\right) > 0\right\}, equals the sum of the residues of the integrand in this half-plane, and find its value.

?
Problem 14.5

Verify the applicability of the residue theorem to the following integrals, taken over the boundaries Γ\Gamma of unbounded domains DD, and compute these integrals:

?
(1)

∫Γze−zz2−1dz(D={z:Re(z)>0})\int_{\Gamma } \frac{z e^{-z}}{z^{2}-1} d z \quad (D = \left\{ z: \mathrm{Re}\left(z\right) > 0\right\} );

(2)

∫Γezsh⁡2zdz(D={z:−π4<Im(z)<π4})\int_{\Gamma } \frac{e^{z}}{\operatorname {sh} 2 z} d z \quad \left(D = \left\{ z:-\frac{\pi }{4} < \mathrm{Im}\left(z\right) < \frac{\pi }{4}\right\} \right);

(3)

∫Γz3(z−1)2e−z3dz(D={z:−π6<arg⁡z<π6})\int_{\Gamma } \frac{z^{3}}{(z-1)^{2}} e^{-z^{3}} d z \quad \left(D = \left\{ z:-\frac{\pi }{6} < \arg z < \frac{\pi }{6}\right\} \right).

Problem 14.6

Verify that the residue theorem is not applicable to the integrals taken over the boundaries Γ\Gamma of the indicated domains DD:

?
(1)

I=∫Γe−z2dz(D={z:Im(z)>0})I = \int_{\Gamma } e^{-z^{2}} d z \quad (D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} );

(2)

I=∫Γsin⁡z1+z2dz(D={z:Im(z)>0})I = \int_{\Gamma } \frac{\sin z}{1+z^{2}} d z \quad (D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} ).

Problem 14.7

Let the functions f(z)f(z) and g(z)g(z) be regular in the closure Dˉ\bar{D} of a bounded domain DD, where the function f(z)f(z) has, in the domain DD, zeros a1,…,ana_{1}, \ldots , a_{n} (each zero written as many times as its order) and has no other zeros either in the domain DD or on its boundary Γ\Gamma. Prove that

∮Γg(z)f′(z)f(z)dz=2πi∑k=1ng(ak) \oint _{\Gamma } g(z) \frac{f^{\prime }(z)}{f(z)} d z = 2 \pi i \sum _{k = 1}^{n} g\left(a_{k}\right)
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Problem 14.8

Let the function f(z)f(z) of Problem 7 also have, in the domain DD, poles b1,…,bmb_{1}, \ldots , b_{m} (each pole also written as many times as its order). Prove that

∮Γg(z)f′(z)f(z)dz=2πi∑k=1ng(ak)−2πi∑k=1mg(bk) \oint _{\Gamma } g(z) \frac{f^{\prime }(z)}{f(z)} d z = 2 \pi i \sum _{k = 1}^{n} g\left(a_{k}\right)-2 \pi i \sum _{k = 1}^{m} g\left(b_{k}\right)
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Problem 14.9

Let the function f(z)f(z) be regular in the closure Dˉ\bar{D} of a bounded domain DD with boundary Γ\Gamma, and let the points a1,…,ana_{1}, \ldots , a_{n} lie in the domain DD and be pairwise distinct. Denote

P(z)=(z−a1)…(z−an) P(z) = \left(z-a_{1}\right) \ldots \left(z-a_{n}\right)

and

Φ(z)=−P(z)2πi∮Γf(ξ)P(ξ)dξξ−z(z∉Dˉ) \Phi (z) = -\frac{P(z)}{2 \pi i} \oint _{\Gamma } \frac{f(\xi )}{P(\xi )} \frac{d \xi }{\xi -z} \quad (z \notin \bar{D})

Prove that the function Φ(z)\Phi (z) extends analytically to the entire plane and is a polynomial of degree n−1n-1 satisfying the conditions

Φ(ak)=f(ak)(k=1,2,…,n). \Phi \left(a_{k}\right) = f\left(a_{k}\right) \quad (k = 1,2, \ldots , n).

(The polynomial Φ(z)\Phi (z) is called the Lagrange interpolation polynomial.)

?
Problem 14.10

Prove the formula

12πi∮∣z∣=1ex2(z+1z)dz=∑n=0∞(x2)2n+11n!(n+1)!. \frac{1}{2 \pi i} \oint _{\left|z\right| = 1} e^{\frac{x}{2}\left(z+\frac{1}{z}\right)} d z = \sum _{n = 0}^{\infty }\left(\frac{x}{2}\right)^{2 n+1} \frac{1}{n!(n+1)!}.
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Section 15The Argument Principle. Rouché's Theorem[15/20%]
Problem 15.1

Find the number of roots of the equations in the domains indicated in parentheses:

?
(1)

z4−3z+1=0({z:∣z∣<1})z^{4}-3 z+1 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} );

(2)

2z4−5z+2=0({z:∣z∣<1})2 z^{4}-5 z+2 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} );

(3)

z7−5z4+z2−2=0({z:∣z∣<1})z^{7}-5 z^{4}+z^{2}-2 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} );

(4)

z8−4z5+z2−1=0({z:∣z∣<1})z^{8}-4 z^{5}+z^{2}-1 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} );

(5)

z3−12z+2=0({z:∣z∣<2})z^{3}-12 z+2 = 0 \quad (\left\{ z:\left|z\right| < 2\right\} );

(6)

z4−9z+1=0({z:∣z∣<2})z^{4}-9 z+1 = 0 \quad (\left\{ z:\left|z\right| < 2\right\} );

(7)

z6−6z+10=0({z:∣z∣>1})z^{6}-6 z+10 = 0 \quad (\left\{ z:\left|z\right| > 1\right\} );

(8)

z4+z3−4z+1=0({z:1<∣z∣<2})z^{4}+z^{3}-4 z+1 = 0 \quad (\left\{ z: 1 < \left|z\right| < 2\right\} );

(9)

z5+z2+3z+1=0({z:∣z∣<1})z^{5}+z^{2}+3 z+1 = 0 \quad (\left\{ z:\left|z\right| < 1\right\} );

(10)

z6+3z4+2z3+1=0({z:∣z∣<2})z^{6}+3 z^{4}+2 z^{3}+1 = 0 \quad (\left\{ z:\left|z\right| < 2\right\} ).

Problem 15.2

Prove that for any complex value aa and any integer n⩾2n \geqslant 2 the equation 1+z+azn=01+z+a z^{n} = 0 has at least one root in the disk {z:∣z∣⩽2}\left\{ z:\left|z\right| \leqslant 2\right\}.

?
Problem 15.3

Prove that for λ>1\lambda > 1 the equation

zeλ−z=1 z e^{\lambda -z} = 1

has exactly one root (which is moreover real) in the disk {z:∣z∣⩽1}\left\{ z:\left|z\right| \leqslant 1\right\}.

?
Problem 15.4

Prove that for λ>1\lambda > 1 the equation

z=λ−e−z z = \lambda -e^{-z}

has exactly one root (which is moreover real) in the half-plane {z:Re(z)⩾0}\left\{ z: \mathrm{Re}\left(z\right) \geqslant 0\right\}.

?
Problem 15.5

Prove that the equation

az3−z+b=e−z(z+2) a z^{3}-z+b = e^{-z}(z+2)

has no roots in the half-plane {z:Re(z)>0}\left\{ z: \mathrm{Re}\left(z\right) > 0\right\} for a>0,b>0a > 0, b > 0.

?
Problem 15.6

Let the function f(z)f(z) be regular in the disk {z:∣z∣<1}\left\{ z:\left|z\right| < 1\right\}. Prove that there exists a number ρ>0\rho > 0 such that for all ww in the disk {w:∣w∣<ρ}\left\{ w:\left|w\right| < \rho \right\} the equation z=wf(z)z = w f(z) has exactly one root in the disk {z:∣z∣<1}\left\{ z:\left|z\right| < 1\right\}.

?
Problem 15.7

Let the function f(z)f(z) be regular in the disk {z:∣z∣<1}\left\{ z:\left|z\right| < 1\right\} with f(0)≠0f(0) \neq 0. Prove that there exists a number ρ>0\rho > 0 such that for all ww in the annulus {w:0<∣w∣<ρ}\left\{ w: 0 < \left|w\right| < \rho \right\} the equation zm=wf(z)z^{m} = w f(z) has exactly mm distinct roots in the disk {z:∣z∣<1}\left\{ z:\left|z\right| < 1\right\}.

?
Problem 15.8

Prove that the equation zsin⁡z=1z \sin z = 1 has only real roots.

?
Problem 15.9

Prove that the equation tg⁡z=z\operatorname {tg} z = z has only real roots.

?
Problem 15.10

Determine the number of roots of the polynomial

P5(z)=z5−12z2+14 P_{5}(z) = z^{5}-12 z^{2}+14

in the right half-plane D={z:Re(z)>0}D = \left\{ z: \mathrm{Re}\left(z\right) > 0\right\}.

?
Problem 15.11

In which quadrants are the roots of the equation z4+z3+4z2+2z+3=0z^{4}+z^{3}+4 z^{2}+2 z+3 = 0 located?

?
Problem 15.12

Prove that if the function f(z)f(z) is regular in a domain DD and for each point zz of DD there exists an nn such that f(n)(z)=0f^{(n)}(z) = 0, then f(z)f(z) is a polynomial.

?
Problem 15.13

Let the functions f(z)f(z) and F(z)F(z) be regular in a bounded domain DD and continuous up to its boundary Γ\Gamma, on which the function Im(f(z)F(z))\mathrm{Im}\left(\frac{f(z)}{F(z)}\right) does not vanish. Prove that the functions F(z)F(z) and F(z)+f(z)F(z)+f(z) have the same number of zeros in the domain DD.

?
Problem 15.14

Let the function f(z)f(z) be regular in a bounded domain DD, except for a finite number of poles, and continuous up to its boundary Γ\Gamma (except at the same poles). Denote

M=max⁡z∈Γ∣f(z)∣. M = \max _{z \in \Gamma }\left|f(z)\right|.

Prove that for every complex value aa satisfying the condition ∣a∣>M\left|a\right| > M, the number of zeros of the function f(z)−af(z)-a in the domain DD equals the number of its poles in this domain.

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Problem 15.15

Let the function f(z)f(z) be regular in a bounded domain DD, except for a finite number of poles, and continuous up to its boundary (except at the same poles). Prove that if the function Im(f(z))\mathrm{Im}\left(f(z)\right) does not vanish on the boundary of the domain DD, then the number of zeros of the function f(z)f(z) in the domain DD equals the number of its poles in this domain.

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