Integral of real part of z around the unit circle












2












$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago
















2












$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago














2












2








2





$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$




question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions




residue-calculus complex-integration analyticity analytic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









AdityaAditya

278314




278314








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago














  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago








2




2




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    40 mins ago



















2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    32 mins ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    30 mins ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    40 mins ago
















3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    40 mins ago














3












3








3





$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$



The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









jmerryjmerry

6,457718




6,457718












  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    40 mins ago


















  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    40 mins ago
















$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
40 mins ago




$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
40 mins ago











2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    32 mins ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    30 mins ago


















2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    32 mins ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    30 mins ago
















2












2








2





$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$



${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









copper.hatcopper.hat

127k559160




127k559160












  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    32 mins ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    30 mins ago




















  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    32 mins ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    30 mins ago


















$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
1 hour ago




$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
1 hour ago












$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
1 hour ago












$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
1 hour ago












$begingroup$
Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
$endgroup$
– Aditya
32 mins ago






$begingroup$
Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
$endgroup$
– Aditya
32 mins ago














$begingroup$
I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
$endgroup$
– copper.hat
30 mins ago






$begingroup$
I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
$endgroup$
– copper.hat
30 mins ago




















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