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
    Feb 2 at 6:19
















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
    Feb 2 at 6:19














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








edited Feb 2 at 9:35









Asaf Karagila♦

305k33435766




305k33435766










asked Feb 2 at 5:49









AdityaAditya

282314




282314








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    Feb 2 at 6:19














  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    Feb 2 at 6:19








2




2




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
Feb 2 at 6:19




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
Feb 2 at 6:19










2 Answers
2






active

oldest

votes


















6












$begingroup$

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
    Feb 2 at 6:40



















3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    Feb 2 at 6:14










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    Feb 2 at 6:14










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    Feb 2 at 6:20






  • 1




    $begingroup$
    Generally I find using the exponential to be more convenient than trigonometric functions.
    $endgroup$
    – copper.hat
    Feb 2 at 6:52






  • 1




    $begingroup$
    It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
    $endgroup$
    – jmerry
    Feb 2 at 10:25











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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
    Feb 2 at 6:40
















6












$begingroup$

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
    Feb 2 at 6:40














6












6








6





$begingroup$

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 Feb 2 at 6:13









jmerryjmerry

10.9k1225




10.9k1225












  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    Feb 2 at 6:40


















  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    Feb 2 at 6:40
















$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
Feb 2 at 6:40




$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
Feb 2 at 6:40











3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    Feb 2 at 6:14










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    Feb 2 at 6:14










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    Feb 2 at 6:20






  • 1




    $begingroup$
    Generally I find using the exponential to be more convenient than trigonometric functions.
    $endgroup$
    – copper.hat
    Feb 2 at 6:52






  • 1




    $begingroup$
    It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
    $endgroup$
    – jmerry
    Feb 2 at 10:25
















3












$begingroup$

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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    Feb 2 at 6:14










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    Feb 2 at 6:14










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    Feb 2 at 6:20






  • 1




    $begingroup$
    Generally I find using the exponential to be more convenient than trigonometric functions.
    $endgroup$
    – copper.hat
    Feb 2 at 6:52






  • 1




    $begingroup$
    It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
    $endgroup$
    – jmerry
    Feb 2 at 10:25














3












3








3





$begingroup$

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






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 Feb 2 at 6:15

























answered Feb 2 at 6:13









copper.hatcopper.hat

127k559160




127k559160












  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    Feb 2 at 6:14










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    Feb 2 at 6:14










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    Feb 2 at 6:20






  • 1




    $begingroup$
    Generally I find using the exponential to be more convenient than trigonometric functions.
    $endgroup$
    – copper.hat
    Feb 2 at 6:52






  • 1




    $begingroup$
    It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
    $endgroup$
    – jmerry
    Feb 2 at 10:25


















  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    Feb 2 at 6:14










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    Feb 2 at 6:14










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    Feb 2 at 6:20






  • 1




    $begingroup$
    Generally I find using the exponential to be more convenient than trigonometric functions.
    $endgroup$
    – copper.hat
    Feb 2 at 6:52






  • 1




    $begingroup$
    It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
    $endgroup$
    – jmerry
    Feb 2 at 10:25
















$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
Feb 2 at 6:14




$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
Feb 2 at 6:14












$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
Feb 2 at 6:14




$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
Feb 2 at 6:14












$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
Feb 2 at 6:20




$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
Feb 2 at 6:20




1




1




$begingroup$
Generally I find using the exponential to be more convenient than trigonometric functions.
$endgroup$
– copper.hat
Feb 2 at 6:52




$begingroup$
Generally I find using the exponential to be more convenient than trigonometric functions.
$endgroup$
– copper.hat
Feb 2 at 6:52




1




1




$begingroup$
It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
$endgroup$
– jmerry
Feb 2 at 10:25




$begingroup$
It doesn't take measure theory - just the definition of a line integral. $int_C f(z),dz$ is defined as $int_a^b f(z(t))cdot z'(t),dt$ for a parametrization $z(t)$ of the curve. When is taking the real part of the integral the same as taking the real part of $f$? When that tangent $z'(t)$ is real - when the curve is just a line parallel to the real axis. Otherwise, the real part of $fcdot z'$ will mix information from real and imaginary parts of $f$.
$endgroup$
– jmerry
Feb 2 at 10:25


















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