Is this true? $Reint{f(z)dz}=int{Re(f(z))dz}$ [closed]
I have to say if this is true or not and why.
Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$
integration complex-analysis
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 at 4:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have to say if this is true or not and why.
Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$
integration complex-analysis
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 at 4:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
2
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
2
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13
add a comment |
I have to say if this is true or not and why.
Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$
integration complex-analysis
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
I have to say if this is true or not and why.
Let f a complex function, then $$Reint_{gamma}{f(z)dz}=int_{gamma}{Re(f(z))dz}$$
integration complex-analysis
integration complex-analysis
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
edited Dec 7 at 23:07
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Dec 7 at 22:13
Juan De Dios Rojas
222
asked Dec 7 at 22:13
Juan De Dios Rojas
222
asked Dec 7 at 22:13
Juan De Dios Rojas
222
222
closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 at 4:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R Dec 8 at 4:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, Henrik, Kavi Rama Murthy, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
2
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
2
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13
add a comment |
2
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
2
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13
2
2
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
2
2
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13
add a comment |
2 Answers
2
active
oldest
votes
$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$
$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$
answered Dec 7 at 23:01
ryszard eggink
308110
add a comment |
$dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.
answered Dec 8 at 0:22
maridia
1,06012
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$
$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$
answered Dec 7 at 23:01
ryszard eggink
308110
add a comment |
$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$
$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$
answered Dec 7 at 23:01
ryszard eggink
308110
add a comment |
$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$
$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$
answered Dec 7 at 23:01
ryszard eggink
308110
$$Reint_{gamma}{frac{i}{z}text{ } dz}=-2pi$$
$$int_{0}^{2pi}-Refrac{1}{e^{it} }dt=int_{0}^{2pi}-costdt=0$$
answered Dec 7 at 23:01
ryszard eggink
308110
answered Dec 7 at 23:01
ryszard eggink
308110
answered Dec 7 at 23:01
ryszard eggink
308110
answered Dec 7 at 23:01
ryszard eggink
308110
308110
add a comment |
add a comment |
$dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.
answered Dec 8 at 0:22
maridia
1,06012
add a comment |
$dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.
answered Dec 8 at 0:22
maridia
1,06012
add a comment |
$dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.
answered Dec 8 at 0:22
maridia
1,06012
$dz$ is complex for a general path $gamma$. Your statement would be true if $gamma$ was on the real line. The point is that for two complex numbers $a$ and $b$, $Re(ab) neq Re(a)b$. Indeed, they are equal iff $b$ is real.
answered Dec 8 at 0:22
maridia
1,06012
answered Dec 8 at 0:22
maridia
1,06012
answered Dec 8 at 0:22
maridia
1,06012
answered Dec 8 at 0:22
maridia
1,06012
1,06012
add a comment |
add a comment |
2
No. ${}{}{}{}{}$
– T. Bongers
Dec 7 at 22:17
2
How about $dz$? Try integrating along an arc of the unit cicle $z=e^{i theta}$.
– mlerma54
Dec 7 at 22:32
Consider that the imaginary part of $f(z)$ could integrate to a real value, or vice versa. It depends a lot on the path $gamma$.
– zahbaz
Dec 7 at 23:13