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







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  • Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    – achille hui
    Oct 5 '13 at 19:27










  • @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    – achille hui
    Oct 5 '13 at 19:34










  • @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).
    – Paul S.
    Oct 5 '13 at 19:44










  • Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    – Paul S.
    Oct 5 '13 at 19:50

















up vote
3
down vote

favorite












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







share|cite|improve this question






















  • Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    – achille hui
    Oct 5 '13 at 19:27










  • @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    – achille hui
    Oct 5 '13 at 19:34










  • @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).
    – Paul S.
    Oct 5 '13 at 19:44










  • Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    – Paul S.
    Oct 5 '13 at 19:50















up vote
3
down vote

favorite









up vote
3
down vote

favorite











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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 5 '13 at 19:15









Paul S.

1,39711539




1,39711539












  • Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    – achille hui
    Oct 5 '13 at 19:27










  • @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    – achille hui
    Oct 5 '13 at 19:34










  • @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).
    – Paul S.
    Oct 5 '13 at 19:44










  • Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    – Paul S.
    Oct 5 '13 at 19:50




















  • Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    – achille hui
    Oct 5 '13 at 19:27










  • @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    – achille hui
    Oct 5 '13 at 19:34










  • @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).
    – Paul S.
    Oct 5 '13 at 19:44










  • Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    – Paul S.
    Oct 5 '13 at 19:50


















Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
– achille hui
Oct 5 '13 at 19:27




Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
– achille hui
Oct 5 '13 at 19:27












@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
– Paul S.
Oct 5 '13 at 19:30




@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
– Paul S.
Oct 5 '13 at 19:30




1




1




$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34




$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34












@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).
– Paul S.
Oct 5 '13 at 19:44




@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).
– Paul S.
Oct 5 '13 at 19:44












Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
– Paul S.
Oct 5 '13 at 19:50






Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
– Paul S.
Oct 5 '13 at 19:50












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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.






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  • 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.
    – Sambo
    Oct 2 '17 at 2:16











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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.






share|cite|improve this answer























  • 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.
    – Sambo
    Oct 2 '17 at 2:16















up vote
0
down vote













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.






share|cite|improve this answer























  • 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.
    – Sambo
    Oct 2 '17 at 2:16













up vote
0
down vote










up vote
0
down vote









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.






share|cite|improve this answer














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.







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edited May 5 '14 at 21:31

























answered May 5 '14 at 19:44









Roman Chokler

82258




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  • 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.
    – Sambo
    Oct 2 '17 at 2:16


















  • 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.
    – Sambo
    Oct 2 '17 at 2:16
















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.
– Sambo
Oct 2 '17 at 2:16




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.
– Sambo
Oct 2 '17 at 2:16


















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