Integral with differential is purely imaginary
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Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
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|
show 1 more comment
$begingroup$
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
$endgroup$
$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
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@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
1
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
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– achille hui
Oct 5 '13 at 19:34
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@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
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Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50
|
show 1 more comment
$begingroup$
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
$endgroup$
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
asked Oct 5 '13 at 19:15
Paul S.Paul S.
1,40211540
1,40211540
$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
1
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34
$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50
|
show 1 more comment
$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
1
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34
$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50
$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
1
1
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34
$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50
$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50
|
show 1 more comment
1 Answer
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How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
$endgroup$
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
add a comment |
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1 Answer
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$begingroup$
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
$endgroup$
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
add a comment |
$begingroup$
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
$endgroup$
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
add a comment |
$begingroup$
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
$endgroup$
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
edited May 5 '14 at 21:31
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
answered May 5 '14 at 19:44
Roman ChoklerRoman Chokler
82258
82258
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
add a comment |
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16
add a comment |
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$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27
$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30
1
$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34
$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44
$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50