Integral of real part of z around the unit circle
$begingroup$
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?
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
$endgroup$
add a comment |
$begingroup$
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?
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
$endgroup$
2
$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago
add a comment |
$begingroup$
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?
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
$endgroup$
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?
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
asked 1 hour ago
AdityaAditya
278314
278314
2
$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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)
$endgroup$
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
add a comment |
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
$begingroup$
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
37 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
36 mins ago
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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)
$endgroup$
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
add a comment |
$begingroup$
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)
$endgroup$
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
add a comment |
$begingroup$
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)
$endgroup$
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)
answered 1 hour ago
jmerryjmerry
6,457718
6,457718
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
add a comment |
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
46 mins ago
add a comment |
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
$begingroup$
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
37 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
36 mins ago
|
show 1 more comment
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
$begingroup$
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
37 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
36 mins ago
|
show 1 more comment
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
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
37 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
36 mins ago
|
show 1 more comment
$begingroup$
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
37 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
36 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
37 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
37 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
36 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
36 mins ago
|
show 1 more comment
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2
$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago