A Strange Mistake in Application of Residue theorem $int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta.$ [closed]

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up vote
3
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$$int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta.$$

While applying the calculus of residues to the above problem I'm getting the answer as
$ 19pi/24$. I have tried many times and rechecked the calculations but getting the same answer.

However in my book the answer given is $pi/6.$

Can someone check and tell me whether I'm correct or the book? Please help.

integration complex-analysis residue-calculus

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edited Aug 2 at 12:08

user 108128

18.8k41544

asked Jul 24 at 13:30

johny

515

closed as off-topic by spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco Jul 25 at 9:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The book is correct.
  – uniquesolution
  Jul 24 at 13:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
  – Adrian Keister
  Jul 24 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't show any work, it is impossible to say where you are making your mistake...
  – paul garrett
  Jul 24 at 16:44
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

$$int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta.$$

While applying the calculus of residues to the above problem I'm getting the answer as
$ 19pi/24$. I have tried many times and rechecked the calculations but getting the same answer.

However in my book the answer given is $pi/6.$

Can someone check and tell me whether I'm correct or the book? Please help.

integration complex-analysis residue-calculus

share|cite|improve this question

edited Aug 2 at 12:08

user 108128

18.8k41544

asked Jul 24 at 13:30

johny

515

closed as off-topic by spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco Jul 25 at 9:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The book is correct.
  – uniquesolution
  Jul 24 at 13:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
  – Adrian Keister
  Jul 24 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't show any work, it is impossible to say where you are making your mistake...
  – paul garrett
  Jul 24 at 16:44
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

$$int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta.$$

While applying the calculus of residues to the above problem I'm getting the answer as
$ 19pi/24$. I have tried many times and rechecked the calculations but getting the same answer.

However in my book the answer given is $pi/6.$

Can someone check and tell me whether I'm correct or the book? Please help.

integration complex-analysis residue-calculus

share|cite|improve this question

edited Aug 2 at 12:08

user 108128

18.8k41544

asked Jul 24 at 13:30

johny

515

$$int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta.$$

While applying the calculus of residues to the above problem I'm getting the answer as
$ 19pi/24$. I have tried many times and rechecked the calculations but getting the same answer.

However in my book the answer given is $pi/6.$

Can someone check and tell me whether I'm correct or the book? Please help.

integration complex-analysis residue-calculus

share|cite|improve this question

edited Aug 2 at 12:08

user 108128

18.8k41544

asked Jul 24 at 13:30

johny

515

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 12:08

user 108128

18.8k41544

edited Aug 2 at 12:08

user 108128

18.8k41544

edited Aug 2 at 12:08

user 108128

18.8k41544

18.8k41544

asked Jul 24 at 13:30

johny

515

asked Jul 24 at 13:30

johny

515

asked Jul 24 at 13:30

johny

515

515

closed as off-topic by spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco Jul 25 at 9:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco Jul 25 at 9:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – spaceisdarkgreen, Claude Leibovici, Jose Arnaldo Bebita Dris, Delta-u, Taroccoesbrocco

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The book is correct.
  – uniquesolution
  Jul 24 at 13:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
  – Adrian Keister
  Jul 24 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't show any work, it is impossible to say where you are making your mistake...
  – paul garrett
  Jul 24 at 16:44
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The book is correct.
  – uniquesolution
  Jul 24 at 13:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
  – Adrian Keister
  Jul 24 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you don't show any work, it is impossible to say where you are making your mistake...
  – paul garrett
  Jul 24 at 16:44
  

  
  
  
  

1

1

The book is correct.
– uniquesolution
Jul 24 at 13:35

The book is correct.
– uniquesolution
Jul 24 at 13:35

1

1

The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
– Adrian Keister
Jul 24 at 13:59

The command Integrate[Cos[2t]/(5+4Cos[t]),t,0,2Pi] in Mathematica shows that the book is correct. Why don't you post some of your working so we can look at it?
– Adrian Keister
Jul 24 at 13:59

If you don't show any work, it is impossible to say where you are making your mistake...
– paul garrett
Jul 24 at 16:44

If you don't show any work, it is impossible to say where you are making your mistake...
– paul garrett
Jul 24 at 16:44

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

If $varphi(x,y)=fracx^2-y^25+4x$, then your integral is$$int_0^2pivarphi(costheta,sintheta),mathrm dtheta.tag1$$Let$$f(z)=frac1zvarphileft(fracz+z^-12,fracz-z^-12iright)$$and$$beginarrayrcccgammacolon&[0,2pi]&longrightarrow&mathbb C\&theta&mapsto&e^itheta.endarray$$Then$$varphi(costheta,sintheta)=varphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)=e^ithetafleft(e^ithetaright).$$Thereforebeginalign(1)&=frac1iint_0^2pie^-ithetavarphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)ie^itheta,mathrm dtheta\&=-iint_gamma f(z),mathrm dz.endalignThis integral can be computed through the residue theorem: it is equal to$$2pisumoperatornameres_z=chibigl(f(z)bigr),tag2$$where the possible values of $chi$ are the poles of $f$ in the open unit circle. But$$f(z)=fracz^4+1z^2(4z^2+10z+4).$$So, $f$ has $3$ poles, two of which are located at the open unit circle: $0$ and $-frac12$ (the third one is $-2$) and the residue of $f$ at these points is $frac1724$ and $-frac58$ respectively. Therefore, your integral is equal to$$2pileft(frac1724-frac58right)=fracpi6.$$

share|cite|improve this answer

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

add a comment | 

up vote
4
down vote

Use substitutions $z=e^itheta$ and $dtheta=dfracdziz$ then
beginalign
int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta
&= bf Reint_dfracz^25+4(z+1/z)dfracdziz \
&= bf Redfrac1iint_dfracfracz^22(z+2)z+1/2dz \
&= bf Redfrac1i2pi idfrac(-frac12)^23 \
&= colorbluedfracpi6
endalign

share|cite|improve this answer

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
  – William
  Aug 2 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's real part.
  – user 108128
  Aug 2 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
  – William
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
  – user 108128
  2 days ago
  
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

If $varphi(x,y)=fracx^2-y^25+4x$, then your integral is$$int_0^2pivarphi(costheta,sintheta),mathrm dtheta.tag1$$Let$$f(z)=frac1zvarphileft(fracz+z^-12,fracz-z^-12iright)$$and$$beginarrayrcccgammacolon&[0,2pi]&longrightarrow&mathbb C\&theta&mapsto&e^itheta.endarray$$Then$$varphi(costheta,sintheta)=varphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)=e^ithetafleft(e^ithetaright).$$Thereforebeginalign(1)&=frac1iint_0^2pie^-ithetavarphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)ie^itheta,mathrm dtheta\&=-iint_gamma f(z),mathrm dz.endalignThis integral can be computed through the residue theorem: it is equal to$$2pisumoperatornameres_z=chibigl(f(z)bigr),tag2$$where the possible values of $chi$ are the poles of $f$ in the open unit circle. But$$f(z)=fracz^4+1z^2(4z^2+10z+4).$$So, $f$ has $3$ poles, two of which are located at the open unit circle: $0$ and $-frac12$ (the third one is $-2$) and the residue of $f$ at these points is $frac1724$ and $-frac58$ respectively. Therefore, your integral is equal to$$2pileft(frac1724-frac58right)=fracpi6.$$

share|cite|improve this answer

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

add a comment | 

up vote
3
down vote



accepted

If $varphi(x,y)=fracx^2-y^25+4x$, then your integral is$$int_0^2pivarphi(costheta,sintheta),mathrm dtheta.tag1$$Let$$f(z)=frac1zvarphileft(fracz+z^-12,fracz-z^-12iright)$$and$$beginarrayrcccgammacolon&[0,2pi]&longrightarrow&mathbb C\&theta&mapsto&e^itheta.endarray$$Then$$varphi(costheta,sintheta)=varphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)=e^ithetafleft(e^ithetaright).$$Thereforebeginalign(1)&=frac1iint_0^2pie^-ithetavarphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)ie^itheta,mathrm dtheta\&=-iint_gamma f(z),mathrm dz.endalignThis integral can be computed through the residue theorem: it is equal to$$2pisumoperatornameres_z=chibigl(f(z)bigr),tag2$$where the possible values of $chi$ are the poles of $f$ in the open unit circle. But$$f(z)=fracz^4+1z^2(4z^2+10z+4).$$So, $f$ has $3$ poles, two of which are located at the open unit circle: $0$ and $-frac12$ (the third one is $-2$) and the residue of $f$ at these points is $frac1724$ and $-frac58$ respectively. Therefore, your integral is equal to$$2pileft(frac1724-frac58right)=fracpi6.$$

share|cite|improve this answer

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

If $varphi(x,y)=fracx^2-y^25+4x$, then your integral is$$int_0^2pivarphi(costheta,sintheta),mathrm dtheta.tag1$$Let$$f(z)=frac1zvarphileft(fracz+z^-12,fracz-z^-12iright)$$and$$beginarrayrcccgammacolon&[0,2pi]&longrightarrow&mathbb C\&theta&mapsto&e^itheta.endarray$$Then$$varphi(costheta,sintheta)=varphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)=e^ithetafleft(e^ithetaright).$$Thereforebeginalign(1)&=frac1iint_0^2pie^-ithetavarphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)ie^itheta,mathrm dtheta\&=-iint_gamma f(z),mathrm dz.endalignThis integral can be computed through the residue theorem: it is equal to$$2pisumoperatornameres_z=chibigl(f(z)bigr),tag2$$where the possible values of $chi$ are the poles of $f$ in the open unit circle. But$$f(z)=fracz^4+1z^2(4z^2+10z+4).$$So, $f$ has $3$ poles, two of which are located at the open unit circle: $0$ and $-frac12$ (the third one is $-2$) and the residue of $f$ at these points is $frac1724$ and $-frac58$ respectively. Therefore, your integral is equal to$$2pileft(frac1724-frac58right)=fracpi6.$$

share|cite|improve this answer

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

If $varphi(x,y)=fracx^2-y^25+4x$, then your integral is$$int_0^2pivarphi(costheta,sintheta),mathrm dtheta.tag1$$Let$$f(z)=frac1zvarphileft(fracz+z^-12,fracz-z^-12iright)$$and$$beginarrayrcccgammacolon&[0,2pi]&longrightarrow&mathbb C\&theta&mapsto&e^itheta.endarray$$Then$$varphi(costheta,sintheta)=varphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)=e^ithetafleft(e^ithetaright).$$Thereforebeginalign(1)&=frac1iint_0^2pie^-ithetavarphileft(frace^itheta+e^-itheta2,frace^itheta-e^-itheta2iright)ie^itheta,mathrm dtheta\&=-iint_gamma f(z),mathrm dz.endalignThis integral can be computed through the residue theorem: it is equal to$$2pisumoperatornameres_z=chibigl(f(z)bigr),tag2$$where the possible values of $chi$ are the poles of $f$ in the open unit circle. But$$f(z)=fracz^4+1z^2(4z^2+10z+4).$$So, $f$ has $3$ poles, two of which are located at the open unit circle: $0$ and $-frac12$ (the third one is $-2$) and the residue of $f$ at these points is $frac1724$ and $-frac58$ respectively. Therefore, your integral is equal to$$2pileft(frac1724-frac58right)=fracpi6.$$

share|cite|improve this answer

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 14:11

edited Jul 24 at 14:11

edited Jul 24 at 14:11

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

answered Jul 24 at 13:58

José Carlos Santos

112k1696172

112k1696172

add a comment | 

add a comment | 

up vote
4
down vote

Use substitutions $z=e^itheta$ and $dtheta=dfracdziz$ then
beginalign
int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta
&= bf Reint_dfracz^25+4(z+1/z)dfracdziz \
&= bf Redfrac1iint_dfracfracz^22(z+2)z+1/2dz \
&= bf Redfrac1i2pi idfrac(-frac12)^23 \
&= colorbluedfracpi6
endalign

share|cite|improve this answer

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
  – William
  Aug 2 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's real part.
  – user 108128
  Aug 2 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
  – William
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
  – user 108128
  2 days ago
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

Use substitutions $z=e^itheta$ and $dtheta=dfracdziz$ then
beginalign
int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta
&= bf Reint_dfracz^25+4(z+1/z)dfracdziz \
&= bf Redfrac1iint_dfracfracz^22(z+2)z+1/2dz \
&= bf Redfrac1i2pi idfrac(-frac12)^23 \
&= colorbluedfracpi6
endalign

share|cite|improve this answer

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
  – William
  Aug 2 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's real part.
  – user 108128
  Aug 2 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
  – William
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
  – user 108128
  2 days ago
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Use substitutions $z=e^itheta$ and $dtheta=dfracdziz$ then
beginalign
int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta
&= bf Reint_dfracz^25+4(z+1/z)dfracdziz \
&= bf Redfrac1iint_dfracfracz^22(z+2)z+1/2dz \
&= bf Redfrac1i2pi idfrac(-frac12)^23 \
&= colorbluedfracpi6
endalign

share|cite|improve this answer

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

Use substitutions $z=e^itheta$ and $dtheta=dfracdziz$ then
beginalign
int_0^2pifrac cos(2theta) 5+4cos(theta), dtheta
&= bf Reint_dfracz^25+4(z+1/z)dfracdziz \
&= bf Redfrac1iint_dfracfracz^22(z+2)z+1/2dz \
&= bf Redfrac1i2pi idfrac(-frac12)^23 \
&= colorbluedfracpi6
endalign

share|cite|improve this answer

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 14:24

edited Jul 24 at 14:24

edited Jul 24 at 14:24

answered Jul 24 at 14:13

user 108128

18.8k41544

answered Jul 24 at 14:13

user 108128

18.8k41544

answered Jul 24 at 14:13

user 108128

18.8k41544

18.8k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
  – William
  Aug 2 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's real part.
  – user 108128
  Aug 2 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
  – William
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
  – user 108128
  2 days ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
  – William
  Aug 2 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's real part.
  – user 108128
  Aug 2 at 13:34
  

  
  
  
  


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  sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
  – William
  2 days ago
  

  
  
  
  


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  It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
  – user 108128
  2 days ago
  
  

  
  
  
  

By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
– William
Aug 2 at 13:33

By "Re" do you mean Res(.) ? I always read Re as the real part. So just clearing things up
– William
Aug 2 at 13:33

It's real part.
– user 108128
Aug 2 at 13:34

It's real part.
– user 108128
Aug 2 at 13:34

sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
– William
2 days ago

sorry for the trouble again, but shouldn't the numerator be $z^2 - i sin theta$?
– William
2 days ago

1

1

It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
– user 108128
2 days ago

It is a pleasure. $cos2theta=bf Re e^2itheta$ and after substitution $e^itheta=z$ will be $cos2theta=bf Rez^2$. @William Also note that $bf Re z^2-isintheta=bf Re z^2$ .
– user 108128
2 days ago

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