If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?
$begingroup$
If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?
This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.
Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...
complex-analysis complex-integration
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
$endgroup$
add a comment |
$begingroup$
If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?
This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.
Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...
complex-analysis complex-integration
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
$endgroup$
5
$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29
add a comment |
$begingroup$
If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?
This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.
Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...
complex-analysis complex-integration
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
$endgroup$
If $intlimits_gamma f(z)dzinBbb R$ does it imply $f(z)in Bbb R~forall zingamma$?
This is from a proof where $gamma$ is a circle and we have $1={1over 2pi}intlimits_gamma f(z)dz={1over 2pi}intlimits_gamma Re~f(z)dz$ and I'm unsure how the second equality is justified.
Clearily $0=intlimits_0^{2pi}ie^{it}dt$ and $ie^{it}notin Bbb R$...
complex-analysis complex-integration
complex-analysis complex-integration
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
edited Dec 31 '18 at 13:40
John Cataldo
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
asked Dec 31 '18 at 13:38
John CataldoJohn Cataldo
1,1371316
1,1371316
5
$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29
add a comment |
5
$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29
5
5
$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$
Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$
Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$
Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
add a comment |
$begingroup$
A you talking about a curve Integral?
If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:
$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
answered Dec 31 '18 at 13:49
A. PA. P
1085
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$
Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$
Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$
Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
add a comment |
$begingroup$
$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$
Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$
Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$
Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
add a comment |
$begingroup$
$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$
Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$
Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$
Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
$endgroup$
$$frac1{2pi} int_gamma f(z) mathrm dz = 1$$
Comparing real parts:
$$Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = 1$$
Taking real parts commutes with everything:
$$1 = Re left( frac1{2pi} int_gamma f(z) mathrm dz right) = frac1{2pi} Re left( int_gamma f(z) mathrm dz right) = frac1{2pi} int_gamma Re left( f(z) right) mathrm dz$$
Notice, though, that $displaystyle int_gamma Re left( f(z) right) mathrm dz = int_gamma f(z) mathrm dz$ does not imply $Re left( f(z) right) = f(z)$.
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:41
Kenny LauKenny Lau
19.9k2159
19.9k2159
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
add a comment |
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
$begingroup$
That's not justification, that's just saying it works because it works
$endgroup$
– Jakobian
Dec 31 '18 at 13:44
1
1
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
$begingroup$
@Jakobian but that's pointing out the author's mistake in thinking that we somehow deduced $f(z) in Bbb R$ when writing down that equality.
$endgroup$
– Kenny Lau
Dec 31 '18 at 13:44
add a comment |
$begingroup$
A you talking about a curve Integral?
If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:
$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
answered Dec 31 '18 at 13:49
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
A you talking about a curve Integral?
If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:
$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
answered Dec 31 '18 at 13:49
A. PA. P
1085
$endgroup$
add a comment |
$begingroup$
A you talking about a curve Integral?
If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:
$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
answered Dec 31 '18 at 13:49
A. PA. P
1085
$endgroup$
A you talking about a curve Integral?
If yes then f is probably a One-Form:
Integration over a curve $gamma : [a,b] to mathbb{R}^n$ is defined as follows:
$$
int_gamma f(z) dz=int_a^b f(gamma(z))gamma'(z) dz
$$
(I know this should be a comment rather than an answer but i don't have sufficient points jet)
answered Dec 31 '18 at 13:49
A. PA. P
1085
edited Dec 31 '18 at 13:59
answered Dec 31 '18 at 13:49
A. PA. P
1085
answered Dec 31 '18 at 13:49
A. PA. P
1085
answered Dec 31 '18 at 13:49
A. PA. P
1085
1085
add a comment |
add a comment |
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The answer is no. For example, if $gamma$ is a closed path and $f(z)$ is holomorphic, then the integral is $0$, and $0 in Bbb R$. This does not mean that $f$ is real on $gamma$.
$endgroup$
– Crostul
Dec 31 '18 at 13:52
$begingroup$
Further specialized example. $f(z) = i$ the constant, and $gamma$ is the unit circle. Then $int_gamma f(z);dz = 0$ but $f(z)$ is never real.
$endgroup$
– GEdgar
Dec 31 '18 at 13:54
$begingroup$
Also $int_gamma f(z) dz in mathbb{C}setminusmathbb{R}$ does not imply that there is a $z$ with $f(z) in mathbb{C}$. Take $f equiv 1$ and $gamma: [0, pi] to mathbb{C}$ with $gamma(t) := e^{it}$. (Naively one could see curve integration as just summation of the function values over the curve, but this severely fails!)
$endgroup$
– ComFreek
Dec 31 '18 at 14:29