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
Feb 2 at 6:19
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
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
add a comment |
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
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
Feb 2 at 6:40
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
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
|
show 4 more comments
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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
Feb 2 at 6:40
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
Feb 2 at 6:40
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 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
add a comment |
$begingroup$
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
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
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
|
show 4 more comments
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
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
|
show 4 more comments
$begingroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
$endgroup$
${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.
edited 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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
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2
$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
Feb 2 at 6:19