Residue at infinity

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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)}.

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.

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}.

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}}.

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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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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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.

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Compute the integrals:

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(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)}.

Let f(z)f(z) be a regular branch of the multivalued function Ln⁡z+13−z\operatorname {Ln} \frac{z+1}{3-z} in the plane with a cut along the segment [−1,3][-1,3], such that f(1+0i)=0f(1+0 i) = 0. Compute the integral

∮∣z∣=7z3f(z)z−2dz \oint _{\left|z\right| = 7} \frac{z^{3} f(z)}{z-2} d z
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Let f(z)f(z) be a regular branch of the multivalued function {z3(6−z)4}\left\{ \sqrt[4]{z^{3}(6-z)}\right\} in the plane with a cut along the segment [0,6][0,6], such that f(3+0i)=3f(3+0 i) = 3. Compute the integral

∮∣z∣=7zf(z)z−3dz \oint _{\left|z\right| = 7} \frac{z f(z)}{z-3} d z
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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡5−z3+z\operatorname {Ln} \frac{5-z}{3+z} in the plane with a cut along the segment [−3,5][-3,5], such that f(1+0i)=0f(1+0 i) = 0. Compute the integral

∮∣z∣=7(z+z2)f(z)z−3dz \oint _{\left|z\right| = 7} \frac{\left(z+z^{2}\right) f(z)}{z-3} d z
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Let f(z)f(z) be a regular branch of the multivalued function {z2(4−z)3}\left\{ \sqrt[3]{z^{2}(4-z)}\right\} in the plane with a cut along the segment [0,4][0,4], such that f(2+0i)=2f(2+0 i) = 2. Compute the integral

∮∣z∣=7zf(z)z−2dz. \oint _{\left|z\right| = 7} \frac{z f(z)}{z-2} d z.
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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡3+izz−1\operatorname {Ln} \frac{3+i z}{z-1} in the plane with a cut along the segment [1,3i][1,3 i], such that f(0)=ln⁡3−πif(0) = \ln 3-\pi i. Compute f(∞)f(\infty ) and the integral

∮∣z∣=4z2f(z)z+3dz. \oint _{\left|z\right| = 4} \frac{z^{2} f(z)}{z+3} d z.
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Let f(z)f(z) be a regular branch of the multivalued function {z3(2i−z)4}\left\{ \sqrt[4]{z^{3}(2 i-z)}\right\} in the plane with a cut along the segment [0,2i][0,2 i], such that

f(2)=228e3πi/16 f(2) = 2 \sqrt[8]{2} e^{3 \pi i / 16}

Compute f(i+0)f(i+0) and the integral

∮∣z∣=3zf(z)z+idz \oint _{\left|z\right| = 3} \frac{z f(z)}{z+i} d z
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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡z+i2−z\operatorname {Ln} \frac{z+i}{2-z} in the plane with a cut along the segment [−i,2][-i, 2], such that f(0)=ln⁡12+52πif(0) = \ln \frac{1}{2}+\frac{5}{2} \pi i. Compute f(∞)f(\infty ) and the integral

∮∣z∣=3z2f(z)z+2dz \oint _{\left|z\right| = 3} \frac{z^{2} f(z)}{z+2} d z
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Let f(z)f(z) be a regular branch of the multivalued function {z2(i−z)3}\left\{ \sqrt[3]{z^{2}(i-z)}\right\} in the plane cut along the segment [0,i][0, i] such that f(1)=26eiπ/4f(1) = \sqrt[6]{2} e^{i \pi / 4}. Compute f(i2+0)f\left(\frac{i}{2}+0\right) and the integral

∮∣z∣=3zf(z)3z+idz \oint _{\left|z\right| = 3} \frac{z f(z)}{3 z+i} d z
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Let g(z)g(z) be a regular branch of the multivalued function Ln⁡2i−zz+1\operatorname {Ln} \frac{2 i-z}{z+1} in the plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z∣=2,−π⩽arg⁡z⩽π2},γ2={z:z=x,−2⩽x⩽−1} \begin{aligned} & \gamma _{1} = \left\{ z:\left|z\right| = 2,-\pi \leqslant \arg z \leqslant \frac{\pi }{2}\right\} , \\ & \gamma _{2} = \left\{ z: z = x,-2 \leqslant x \leqslant -1\right\} \end{aligned}

such that g(0)=ln⁡2−3πi2g(0) = \ln 2-\frac{3 \pi i}{2}. Compute the integral

∮z=4zg(z)1+tg⁡1zdz \oint _{z = 4} \frac{z g(z)}{1+\operatorname {tg} \frac{1}{z}} d z
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Let f(z)f(z) be a regular branch of the multivalued function {z2(i−z)3}\left\{ \sqrt[3]{z^{2}(i-z)}\right\} in the plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z+i2∣=32,Re(z)⩾0}γ2={z:∣z+i∣=1,Re(z)⩽0} \begin{aligned} & \gamma _{1} = \left\{ z:\left|z+\frac{i}{2}\right| = \frac{3}{2}, \mathrm{Re}\left(z\right) \geqslant 0\right\} \\ & \gamma _{2} = \left\{ z:\left|z+i\right| = 1, \mathrm{Re}\left(z\right) \leqslant 0\right\} \end{aligned}

such that f(−i)=23ei7π/6f(-i) = \sqrt[3]{2} e^{i 7 \pi / 6}. Compute the integral

∮∣z∣=4f(z)1+e2/zdz \oint _{\left|z\right| = 4} \frac{f(z)}{1+e^{2 / z}} d z
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Let g(z)g(z) be a regular branch of the multivalued function Ln⁡1−ziz+1\operatorname {Ln} \frac{1-z}{i z+1} in the plane cut along the curve

γ={z:∣z∣=1,π2⩽arg⁡z⩽2π} \gamma = \left\{ z:\left|z\right| = 1, \frac{\pi }{2} \leqslant \arg z \leqslant 2 \pi \right\}

such that g(0)=−4πig(0) = -4 \pi i. Compute the integral

∮∣z∣=5zg(z)sin⁡1z+cos⁡1zdz. \oint _{\left|z\right| = 5} \frac{z g(z)}{\sin \frac{1}{z}+\cos \frac{1}{z}} d z.
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Let f(z)f(z) be a regular branch of the multivalued function {z2(2i+z)24}\left\{ \sqrt[4]{z^{2}(2 i+z)^{2}}\right\} in the plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z+2i∣=2,Re(z)⩽0},γ2={z:∣z+3i∣=1,Re(z)⩾0}, \begin{aligned} & \gamma _{1} = \left\{ z:\left|z+2 i\right| = 2, \mathrm{Re}\left(z\right) \leqslant 0\right\} , \\ & \gamma _{2} = \left\{ z:\left|z+3 i\right| = 1, \mathrm{Re}\left(z\right) \geqslant 0\right\} , \end{aligned}

such that f(−3i)=3eπif(-3 i) = \sqrt{3} e^{\pi i}. Compute the integral

∮∣z∣=5f(z)1+2sin⁡1zdz. \oint _{\left|z\right| = 5} \frac{f(z)}{1+2 \sin \frac{1}{z}} d z.
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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡2+iz2+z\operatorname {Ln} \frac{2+i z}{2+z} in the plane cut along the curve

γ={z:∣z∣=2,−π⩽arg⁡z⩽π2} \gamma = \left\{ z:\left|z\right| = 2,-\pi \leqslant \arg z \leqslant \frac{\pi }{2}\right\}

such that h(∞)=12πih(\infty ) = \frac{1}{2} \pi i. Also compute the integral

∮∣z∣=1ezh(z)sin⁡3zdz \oint _{\left|z\right| = 1} \frac{e^{z} h(z)}{\sin ^{3} z} d z
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Let g(z)g(z) be a regular branch of the multivalued function {2z2+1}\left\{ \sqrt{2 z^{2}+1}\right\} in the plane cut along the curve γ={z:∣z∣=12,Re(z)⩾0}\gamma = \left\{ z:\left|z\right| = \frac{1}{\sqrt{2}}, \mathrm{Re}\left(z\right) \geqslant 0\right\}, where g(0)=1g(0) = 1. Compute the integral

∮∣z∣=1dz(z−2)(g(z)+3) \oint _{\left|z\right| = 1} \frac{d z}{(z-2)(g(z)+3)}
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Let g(z)g(z) be a regular branch of the multivalued function {(z−2)2(2i−z)3}\left\{ \sqrt[3]{(z-2)^{2}(2 i-z)}\right\} in the plane cut along the segment [2i,2][2 i, 2] such that the principal part of its Laurent series at ∞\infty equals e−iπ/3ze^{-i \pi / 3} z. Compute the integral

∮∣z∣=1g(z)sh⁡3zdz. \oint _{\left|z\right| = 1} \frac{g(z)}{\operatorname {sh}^{3} z} d z.
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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡3+ziz−3\operatorname {Ln} \frac{3+z}{i z-3} in the plane cut along the curve

γ={z:∣z∣=3,−π2⩽arg⁡z⩽π} \gamma = \left\{ z:\left|z\right| = 3,-\frac{\pi }{2} \leqslant \arg z \leqslant \pi \right\}

such that h(∞)=−52πih(\infty ) = -\frac{5}{2} \pi i. Compute the integral

∮∣z∣=1dz(h2(z)+π2)2. \oint _{\left|z\right| = 1} \frac{d z}{\left(h^{2}(z)+\pi ^{2}\right)^{2}}.
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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡2+ziz−1\operatorname {Ln} \frac{2+z}{i z-1} in the plane cut along the curve

γ={z:∣z∣=1,−π2⩽arg⁡z⩽π}∪[−2,−1] \gamma = \left\{ z:\left|z\right| = 1,-\frac{\pi }{2} \leqslant \arg z \leqslant \pi \right\} \cup [-2,-1]

such that Im(h(∞))=3π2\mathrm{Im}\left(h(\infty )\right) = \frac{3 \pi }{2}. Find h(0)h(0) and compute the integral

∮∣z∣=12h(z)sh⁡3zdz. \oint _{\left|z\right| = \frac{1}{2}} \frac{h(z)}{\operatorname {sh}^{3} z} d z.
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Let g(z)g(z) be a regular branch of the multivalued function {1+2z2}\left\{ \sqrt{1+2 z^{2}}\right\} in the plane cut along the curve

γ={z:∣z∣=12,−π2⩽arg⁡z⩽π} \gamma = \left\{ z:\left|z\right| = \frac{1}{\sqrt{2}},-\frac{\pi }{2} \leqslant \arg z \leqslant \pi \right\}

such that g(0)=1g(0) = 1. Let f(z)=z(g(z)+3)2f(z) = \frac{z}{(g(z)+3)^{2}}. Find resz=∞f\mathop{\mathrm{res}}_{z = \infty } f and compute the integral

∮∣z∣=1f(z)dz \oint _{\left|z\right| = 1} f(z) d z
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Let g(z)g(z) be a regular branch of the multivalued function {z2+13}\left\{ \sqrt[3]{z^{2}+1}\right\} in the complex plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z∣=1,0⩽arg⁡z⩽3π2},γ2={z:Im(z)=−1,Re(z)⩾0} \begin{aligned} & \gamma _{1} = \left\{ z:\left|z\right| = 1,0 \leqslant \arg z \leqslant \frac{3 \pi }{2}\right\} , \\ & \gamma _{2} = \left\{ z: \mathrm{Im}\left(z\right) = -1, \mathrm{Re}\left(z\right) \geqslant 0\right\} \end{aligned}

with g(0)=e−2πi/3g(0) = e^{-2 \pi i / 3}. Compute the integral

∮∣z+3∣=1(g(z)g(z)−2)2dz \oint _{\left|z+3\right| = 1}\left(\frac{g(z)}{g(z)-2}\right)^{2} d z
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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(z2−1)\operatorname {Ln}\left(z^{2}-1\right) in the complex plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z∣=1,−π⩽arg⁡z⩽0},γ2={z:Im(z)=0,Re(z)⩾1}, \begin{aligned} & \gamma _{1} = \left\{ z:\left|z\right| = 1,-\pi \leqslant \arg z \leqslant 0\right\} , \\ & \gamma _{2} = \left\{ z: \mathrm{Im}\left(z\right) = 0, \mathrm{Re}\left(z\right) \geqslant 1\right\} , \end{aligned}

with Im(h(−2i))=3π\mathrm{Im}\left(h(-2 i)\right) = 3 \pi. Compute the integral

∮∣z−3i∣=22/7z(h(z)h(z)−πi)2dz \oint _{\left|z-3 i\right| = 22 / 7} z\left(\frac{h(z)}{h(z)-\pi i}\right)^{2} d z
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Let g(z)g(z) be a regular branch of the multivalued function {z2−14}\left\{ \sqrt[4]{z^{2}-1}\right\} in the complex plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z∣=1,π⩽arg⁡z⩽2π}γ2={z:Im(z)⩽−1,Re(z)=0} \begin{aligned} & \gamma _{1} = \left\{ z:\left|z\right| = 1, \pi \leqslant \arg z \leqslant 2 \pi \right\} \\ & \gamma _{2} = \left\{ z: \mathrm{Im}\left(z\right) \leqslant -1, \mathrm{Re}\left(z\right) = 0\right\} \end{aligned}

with g(2)=34g(2) = \sqrt[4]{3}. Compute the integral

∮∣z−i∣=54z(g(z)g(z)−eπi/4)2dz. \oint _{\left|z-i\right| = \frac{5}{4}} z\left(\frac{g(z)}{g(z)-e^{\pi i / 4}}\right)^{2} d z.
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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡(z2+1)\operatorname {Ln}\left(z^{2}+1\right) in the complex plane cut along the curve γ=γ1∪γ2\gamma = \gamma_{1} \cup \gamma_{2}, where

γ1={z:∣z∣=1,π2⩽arg⁡z⩽3π2},γ2={z:Im(z)=1,Re(z)⩾0}, \begin{aligned} & \gamma _{1} = \left\{ z:\left|z\right| = 1, \frac{\pi }{2} \leqslant \arg z \leqslant \frac{3 \pi }{2}\right\} , \\ & \gamma _{2} = \left\{ z: \mathrm{Im}\left(z\right) = 1, \mathrm{Re}\left(z\right) \geqslant 0\right\} , \end{aligned}

with Im(h(0))=0\mathrm{Im}\left(h(0)\right) = 0. Compute the integral

∮∣z+3i∣=74(h(z)h(z)+πi)2dz \oint _{\left|z+3 i\right| = \frac{7}{4}}\left(\frac{h(z)}{h(z)+\pi i}\right)^{2} d z
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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(z−1z+1)\operatorname {Ln}\left(\frac{z-1}{z+1}\right) in the domain G={z:∣z∣>1}G = \left\{ z:\left|z\right| > 1\right\} such that f(∞)=0f(\infty ) = 0. Prove that the multivalued function {f(z)+4i}\left\{ \sqrt{f(z)+4 i}\right\} splits in GG into regular branches.

Let g(z)g(z) be a regular branch of the function {f(z)+4i}\left\{ \sqrt{f(z)+4 i}\right\} in GG such that g(∞)=−2(i+1)g(\infty ) = -\sqrt{2}(i+1). Compute the integral ∮∣z∣=2dzg(z)\oint_{\left|z\right| = 2} \frac{d z}{g(z)}.

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Let f(z)f(z) be a regular branch of the multivalued function {3−z3+z}\left\{ \sqrt{\frac{3-z}{3+z}}\right\} in the domain G={z:∣z∣>3}G = \left\{ z:\left|z\right| > 3\right\} such that f(∞)=−if(\infty ) = -i. Prove that the multivalued function Ln⁡(f(z)−i)\operatorname {Ln}(f(z)-i) splits in GG into regular branches.

Let g(z)g(z) be a regular branch of the function Ln⁡(f(z)−i)\operatorname {Ln}(f(z)-i) in GG such that g(∞)=ln⁡2−iπ2g(\infty ) = \ln 2-i \frac{\pi }{2}. Compute the integral ∮∣z∣=4dzg(z)\oint_{\left|z\right| = 4} \frac{d z}{g(z)}.

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Compute the integrals:

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(1)

∫−22dx(2+x)2(4−x2)4\int_{-2}^{2} \frac{d x}{\sqrt[4]{(2+x)^{2}\left(4-x^{2}\right)}};

(2)

∫02dxx2(2−x)3\int_{0}^{2} \frac{d x}{\sqrt[3]{x^{2}(2-x)}};

(3)

∫01dx(x−3)2x−x2\int_{0}^{1} \frac{d x}{(x-3)^{2} \sqrt{x-x^{2}}};

(4)

∫061x+2x6−xdx\int_{0}^{6} \frac{1}{x+2} \sqrt{\frac{x}{6-x}} d x;

(5)

∫05x(5−x)74x2−5x−6dx\int_{0}^{5} \frac{\sqrt[4]{x(5-x)^{7}}}{x^{2}-5 x-6} d x;

(6)

∫−11x1−x2x2−3x+2dx\int_{-1}^{1} \frac{x \sqrt{1-x^{2}}}{x^{2}-3 x+2} d x;

(7)

∫03dx(x+2)2x3(3−x)4\int_{0}^{3} \frac{d x}{(x+2)^{2} \sqrt[4]{x^{3}(3-x)}};

(8)

∫02x2(2−x)3(x+2)2dx\int_{0}^{2} \frac{\sqrt[3]{x^{2}(2-x)}}{(x+2)^{2}} d x;

(9)

∫03(3−x)x23dx\int_{0}^{3} \sqrt[3]{(3-x) x^{2}} d x;

(10)

∫01x−x220x+39x+2dx\int_{0}^{1} \sqrt{x-x^{2}} \frac{20 x+39}{x+2} d x;

(11)

∫−111+x1−x43x2+13x+20(x+2)2dx\int_{-1}^{1} \sqrt[4]{\frac{1+x}{1-x}} \frac{3 x^{2}+13 x+20}{(x+2)^{2}} d x;

(12)

∫−10−1−1x3x+13x+4dx\int_{-1}^{0} \sqrt{-1-\frac{1}{x}} \frac{3 x+1}{3 x+4} d x.

References

[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.