Evaluate $int_tan z dz$

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Evaluate $int_=4tan z,mathrm dz$

$tan z=fracsin zcos z$ there is a a simple pole at $z=fracpi2$

$operatornameRes(f,fracpi2)=lim_zto fracpi2(z-fracpi2)(fracsin zcos z)$

$operatornameRes(f,frac3pi2)=lim_zto frac3pi2(z-frac3pi2)(fracsin zcos z)$

How to continue ?

complex-analysis contour-integration residue-calculus

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edited Jul 16 at 15:48

José Carlos Santos

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asked Jul 16 at 15:35

newhere

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  Use tan z intead of $tgz$.
  – amWhy
  Jul 16 at 15:39
  

  
  
  
  

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up vote
0
down vote

favorite

Evaluate $int_=4tan z,mathrm dz$

$tan z=fracsin zcos z$ there is a a simple pole at $z=fracpi2$

$operatornameRes(f,fracpi2)=lim_zto fracpi2(z-fracpi2)(fracsin zcos z)$

$operatornameRes(f,frac3pi2)=lim_zto frac3pi2(z-frac3pi2)(fracsin zcos z)$

How to continue ?

complex-analysis contour-integration residue-calculus

share|cite|improve this question

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

asked Jul 16 at 15:35

newhere

759310

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use tan z intead of $tgz$.
  – amWhy
  Jul 16 at 15:39
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Evaluate $int_=4tan z,mathrm dz$

$tan z=fracsin zcos z$ there is a a simple pole at $z=fracpi2$

$operatornameRes(f,fracpi2)=lim_zto fracpi2(z-fracpi2)(fracsin zcos z)$

$operatornameRes(f,frac3pi2)=lim_zto frac3pi2(z-frac3pi2)(fracsin zcos z)$

How to continue ?

complex-analysis contour-integration residue-calculus

share|cite|improve this question

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

asked Jul 16 at 15:35

newhere

759310

Evaluate $int_=4tan z,mathrm dz$

$tan z=fracsin zcos z$ there is a a simple pole at $z=fracpi2$

$operatornameRes(f,fracpi2)=lim_zto fracpi2(z-fracpi2)(fracsin zcos z)$

$operatornameRes(f,frac3pi2)=lim_zto frac3pi2(z-frac3pi2)(fracsin zcos z)$

How to continue ?

complex-analysis contour-integration residue-calculus

share|cite|improve this question

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

asked Jul 16 at 15:35

newhere

759310

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

edited Jul 16 at 15:48

José Carlos Santos

114k1698177

114k1698177

asked Jul 16 at 15:35

newhere

759310

asked Jul 16 at 15:35

newhere

759310

asked Jul 16 at 15:35

newhere

759310

759310

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use tan z intead of $tgz$.
  – amWhy
  Jul 16 at 15:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use tan z intead of $tgz$.
  – amWhy
  Jul 16 at 15:39
  

  
  
  
  

1

1

Use tan z intead of $tgz$.
– amWhy
Jul 16 at 15:39

Use tan z intead of $tgz$.
– amWhy
Jul 16 at 15:39

add a comment | 

2 Answers
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You havebeginalignoperatornameResleft(tan z,pmfracpi2right)&=lim_ztopmfracpi2left(zmpfracpi2right)fracsin zcos z\&=lim_ztopmfracpi2fracsin zfraccos(z)-cosleft(pmfracpi2right)zmpfracpi2\&=fracsinleft(pmfracpi2right)cos'left(pmfracpi2right)\&=-1.endalignBesides, $pmfracpi2$ are the only poles in the disc centered at $0$ with radius $4$. Thereforebeginalignint_=4tan z,mathrm dz&=2pi ileft(operatornameResleft(tan z,fracpi2right)+operatornameResleft(tan z,-fracpi2right)right)\&=-4pi i.endalign

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edited Jul 16 at 16:06

Chris Custer

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

add a comment | 

up vote
2
down vote

Taylor expand $cos(z)=cos(fracpi2)+(z-fracpi 2)cos'(fracpi 2) + frac12(z-fracpi 2)^2 cos''(fracpi 2) + dots$ and $cos(z)=cos(frac3pi2)+(z-frac3pi 2)cos'(frac3pi 2) + frac12(z-frac3pi 2)^2 cos''(frac3pi 2) + dots$, use that to compute the limits, then use the residue theorem.

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answered Jul 16 at 15:41

Kusma

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

You havebeginalignoperatornameResleft(tan z,pmfracpi2right)&=lim_ztopmfracpi2left(zmpfracpi2right)fracsin zcos z\&=lim_ztopmfracpi2fracsin zfraccos(z)-cosleft(pmfracpi2right)zmpfracpi2\&=fracsinleft(pmfracpi2right)cos'left(pmfracpi2right)\&=-1.endalignBesides, $pmfracpi2$ are the only poles in the disc centered at $0$ with radius $4$. Thereforebeginalignint_=4tan z,mathrm dz&=2pi ileft(operatornameResleft(tan z,fracpi2right)+operatornameResleft(tan z,-fracpi2right)right)\&=-4pi i.endalign

share|cite|improve this answer

edited Jul 16 at 16:06

Chris Custer

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

add a comment | 

up vote
5
down vote



accepted

You havebeginalignoperatornameResleft(tan z,pmfracpi2right)&=lim_ztopmfracpi2left(zmpfracpi2right)fracsin zcos z\&=lim_ztopmfracpi2fracsin zfraccos(z)-cosleft(pmfracpi2right)zmpfracpi2\&=fracsinleft(pmfracpi2right)cos'left(pmfracpi2right)\&=-1.endalignBesides, $pmfracpi2$ are the only poles in the disc centered at $0$ with radius $4$. Thereforebeginalignint_=4tan z,mathrm dz&=2pi ileft(operatornameResleft(tan z,fracpi2right)+operatornameResleft(tan z,-fracpi2right)right)\&=-4pi i.endalign

share|cite|improve this answer

edited Jul 16 at 16:06

Chris Custer

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

You havebeginalignoperatornameResleft(tan z,pmfracpi2right)&=lim_ztopmfracpi2left(zmpfracpi2right)fracsin zcos z\&=lim_ztopmfracpi2fracsin zfraccos(z)-cosleft(pmfracpi2right)zmpfracpi2\&=fracsinleft(pmfracpi2right)cos'left(pmfracpi2right)\&=-1.endalignBesides, $pmfracpi2$ are the only poles in the disc centered at $0$ with radius $4$. Thereforebeginalignint_=4tan z,mathrm dz&=2pi ileft(operatornameResleft(tan z,fracpi2right)+operatornameResleft(tan z,-fracpi2right)right)\&=-4pi i.endalign

share|cite|improve this answer

edited Jul 16 at 16:06

Chris Custer

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

You havebeginalignoperatornameResleft(tan z,pmfracpi2right)&=lim_ztopmfracpi2left(zmpfracpi2right)fracsin zcos z\&=lim_ztopmfracpi2fracsin zfraccos(z)-cosleft(pmfracpi2right)zmpfracpi2\&=fracsinleft(pmfracpi2right)cos'left(pmfracpi2right)\&=-1.endalignBesides, $pmfracpi2$ are the only poles in the disc centered at $0$ with radius $4$. Thereforebeginalignint_=4tan z,mathrm dz&=2pi ileft(operatornameResleft(tan z,fracpi2right)+operatornameResleft(tan z,-fracpi2right)right)\&=-4pi i.endalign

share|cite|improve this answer

edited Jul 16 at 16:06

Chris Custer

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 16:06

Chris Custer

5,4582622

edited Jul 16 at 16:06

Chris Custer

5,4582622

edited Jul 16 at 16:06

Chris Custer

5,4582622

5,4582622

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

answered Jul 16 at 15:46

José Carlos Santos

114k1698177

114k1698177

add a comment | 

add a comment | 

up vote
2
down vote

Taylor expand $cos(z)=cos(fracpi2)+(z-fracpi 2)cos'(fracpi 2) + frac12(z-fracpi 2)^2 cos''(fracpi 2) + dots$ and $cos(z)=cos(frac3pi2)+(z-frac3pi 2)cos'(frac3pi 2) + frac12(z-frac3pi 2)^2 cos''(frac3pi 2) + dots$, use that to compute the limits, then use the residue theorem.

share|cite|improve this answer

answered Jul 16 at 15:41

Kusma

1,127112

add a comment | 

up vote
2
down vote

Taylor expand $cos(z)=cos(fracpi2)+(z-fracpi 2)cos'(fracpi 2) + frac12(z-fracpi 2)^2 cos''(fracpi 2) + dots$ and $cos(z)=cos(frac3pi2)+(z-frac3pi 2)cos'(frac3pi 2) + frac12(z-frac3pi 2)^2 cos''(frac3pi 2) + dots$, use that to compute the limits, then use the residue theorem.

share|cite|improve this answer

answered Jul 16 at 15:41

Kusma

1,127112

add a comment | 

up vote
2
down vote

up vote
2
down vote

Taylor expand $cos(z)=cos(fracpi2)+(z-fracpi 2)cos'(fracpi 2) + frac12(z-fracpi 2)^2 cos''(fracpi 2) + dots$ and $cos(z)=cos(frac3pi2)+(z-frac3pi 2)cos'(frac3pi 2) + frac12(z-frac3pi 2)^2 cos''(frac3pi 2) + dots$, use that to compute the limits, then use the residue theorem.

share|cite|improve this answer

answered Jul 16 at 15:41

Kusma

1,127112

Taylor expand $cos(z)=cos(fracpi2)+(z-fracpi 2)cos'(fracpi 2) + frac12(z-fracpi 2)^2 cos''(fracpi 2) + dots$ and $cos(z)=cos(frac3pi2)+(z-frac3pi 2)cos'(frac3pi 2) + frac12(z-frac3pi 2)^2 cos''(frac3pi 2) + dots$, use that to compute the limits, then use the residue theorem.

share|cite|improve this answer

answered Jul 16 at 15:41

Kusma

1,127112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 15:41

Kusma

1,127112

answered Jul 16 at 15:41

Kusma

1,127112

answered Jul 16 at 15:41

Kusma

1,127112

1,127112

add a comment | 

add a comment | 

 

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