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 $int_gammaoverlinef(z)f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverlinef(z(t))f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
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show 1 more comment
up vote
3
down vote
favorite
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverlinef(z)f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverlinef(z(t))f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
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
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
@achillehui It is $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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
 |Â
show 1 more comment
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 $int_gammaoverlinef(z)f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverlinef(z(t))f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverlinef(z)f'(z)dz$ is purely imaginary.
I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverlinef(z(t))f'(z(t))z'(t)dt$. Why will this be purely imaginary?
complex-analysis
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
asked Oct 5 '13 at 19:15
Paul S.
1,36711436
1,36711436
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
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
@achillehui It is $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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
 |Â
show 1 more comment
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
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
@achillehui It is $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
@achillehui It is $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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 $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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
 |Â
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.
$$int_gammaoverlinef(z)f'(z) dz=int_gamma|f(z)|^2fracf'(z)f(z) dz=int_gamma e^f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))e^2ln f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))f(z)f'(z) dz=int_gamma e^-2itheta(z)r(z)e^itheta(z)[r'(z)e^itheta(z)+ir(z)theta'(z)e^itheta(z)] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac12r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
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 Chokler
82258
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
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverlinef(z)f'(z) dz=int_gamma|f(z)|^2fracf'(z)f(z) dz=int_gamma e^f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))e^2ln f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))f(z)f'(z) dz=int_gamma e^-2itheta(z)r(z)e^itheta(z)[r'(z)e^itheta(z)+ir(z)theta'(z)e^itheta(z)] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac12r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
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 Chokler
82258
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
add a comment |Â
up vote
0
down vote
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverlinef(z)f'(z) dz=int_gamma|f(z)|^2fracf'(z)f(z) dz=int_gamma e^f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))e^2ln f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))f(z)f'(z) dz=int_gamma e^-2itheta(z)r(z)e^itheta(z)[r'(z)e^itheta(z)+ir(z)theta'(z)e^itheta(z)] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac12r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
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 Chokler
82258
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
add a comment |Â
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.
$$int_gammaoverlinef(z)f'(z) dz=int_gamma|f(z)|^2fracf'(z)f(z) dz=int_gamma e^f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))e^2ln f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))f(z)f'(z) dz=int_gamma e^-2itheta(z)r(z)e^itheta(z)[r'(z)e^itheta(z)+ir(z)theta'(z)e^itheta(z)] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac12r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
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 Chokler
82258
How about this solution
Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.
$$int_gammaoverlinef(z)f'(z) dz=int_gamma|f(z)|^2fracf'(z)f(z) dz=int_gamma e^f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))e^2ln f(z)fracf'(z)f(z) dz=int_gamma e^-2iarg(f(z))f(z)f'(z) dz=int_gamma e^-2itheta(z)r(z)e^itheta(z)[r'(z)e^itheta(z)+ir(z)theta'(z)e^itheta(z)] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac12r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$
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 Chokler
82258
edited May 5 '14 at 21:31
answered May 5 '14 at 19:44
Roman Chokler
82258
answered May 5 '14 at 19:44
Roman Chokler
82258
answered May 5 '14 at 19:44
Roman Chokler
82258
82258
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
add a comment |Â
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
add a comment |Â
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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
$2^nd$ Hint: what is $fracddt |f(z(t))|^2$?
– achille hui
Oct 5 '13 at 19:34
@achillehui It is $overlinef(z(t))f'(z(t))z'(t)+f(z(t))overlinef'(z(t))overlinez'(t)$. This can be written as $2Re[overlinef(z(t))f'(z(t))z'(t)]$. 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