Integral of complex number over a contour
$begingroup$
I have
$$int_{-1}^1 |z|dz$$
I need to calculate the integral where the integration contour is the upper semi-circle with unit radius. I calculated the integral in $(-1; 1)$ section; the answer is 1, but I'm not sure how to calculate for the upper semi-circle.
contour-integration complex-integration
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
$endgroup$
add a comment |
$begingroup$
I have
$$int_{-1}^1 |z|dz$$
I need to calculate the integral where the integration contour is the upper semi-circle with unit radius. I calculated the integral in $(-1; 1)$ section; the answer is 1, but I'm not sure how to calculate for the upper semi-circle.
contour-integration complex-integration
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
$endgroup$
$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02
add a comment |
$begingroup$
I have
$$int_{-1}^1 |z|dz$$
I need to calculate the integral where the integration contour is the upper semi-circle with unit radius. I calculated the integral in $(-1; 1)$ section; the answer is 1, but I'm not sure how to calculate for the upper semi-circle.
contour-integration complex-integration
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
$endgroup$
I have
$$int_{-1}^1 |z|dz$$
I need to calculate the integral where the integration contour is the upper semi-circle with unit radius. I calculated the integral in $(-1; 1)$ section; the answer is 1, but I'm not sure how to calculate for the upper semi-circle.
contour-integration complex-integration
contour-integration complex-integration
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
edited Nov 28 '18 at 19:02
user3132457
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
asked Nov 28 '18 at 18:53
user3132457user3132457
1336
1336
$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02
add a comment |
$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02
$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note: I think I may have been considering a different contour, since the original poster asked what they could do after finding $intlimits_{-1}^1 |z|dz$. I am thinking of a contour like this: 
where $R=1$. If OP only requires the upper half, then we can disregard the $1$ in the final sum.
If you're taking $gamma$ to be this contour, note that $gamma$ is formed by two curves: $gamma_1$ and $gamma_2$ where $gamma_1$ is the interval $[-1,1]$ and $gamma_2$ is the upper arc of the semicircle. Thus we have that $$ intlimits_gamma |z| dz = intlimits_{gamma_1} |z|dz + intlimits_{gamma_2}|z|dz$$
You already found that $$ intlimits_{gamma_1} |z|dz = intlimits_{-1}^1 |z| dz = 1$$
We have to find $intlimits_{gamma_2}|z|dz$. Note that for all $z$ on $gamma_2$, $|z|=1$ because $z$ lies on the arc of the unit circle. So begin{align*}intlimits_{gamma_2}|z|dz = intlimits_{gamma_2}1dz = intlimits_0^pi ie^{itheta}dtheta = -2
end{align*}
So $$ intlimits_gamma |z| dz = -1$$
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
$endgroup$
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
add a comment |
$begingroup$
Are you integrating clockwise or anti-clockwise?
Assuming you're integrating anti-clockwise, use a substitution of $z=mathrm e^{mathrm i theta}$, where $0 le theta le pi$.
It follows that $mathrm dz = mathrm{ie}^{mathrm i theta}~mathrm dtheta$.
Hence $$int_C |z|~mathrm dz = int_0^{pi} |mathrm e^{mathrm i theta}| ~mathrm{ie}^{mathrm i theta}~mathrm dtheta = mathrm iint_0^{pi}mathrm e^{mathrm itheta}~mathrm dtheta$$
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
$endgroup$
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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oldest
votes
$begingroup$
Note: I think I may have been considering a different contour, since the original poster asked what they could do after finding $intlimits_{-1}^1 |z|dz$. I am thinking of a contour like this:
where $R=1$. If OP only requires the upper half, then we can disregard the $1$ in the final sum.
If you're taking $gamma$ to be this contour, note that $gamma$ is formed by two curves: $gamma_1$ and $gamma_2$ where $gamma_1$ is the interval $[-1,1]$ and $gamma_2$ is the upper arc of the semicircle. Thus we have that $$ intlimits_gamma |z| dz = intlimits_{gamma_1} |z|dz + intlimits_{gamma_2}|z|dz$$
You already found that $$ intlimits_{gamma_1} |z|dz = intlimits_{-1}^1 |z| dz = 1$$
We have to find $intlimits_{gamma_2}|z|dz$. Note that for all $z$ on $gamma_2$, $|z|=1$ because $z$ lies on the arc of the unit circle. So begin{align*}intlimits_{gamma_2}|z|dz = intlimits_{gamma_2}1dz = intlimits_0^pi ie^{itheta}dtheta = -2
end{align*}
So $$ intlimits_gamma |z| dz = -1$$
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
$endgroup$
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
add a comment |
$begingroup$
Note: I think I may have been considering a different contour, since the original poster asked what they could do after finding $intlimits_{-1}^1 |z|dz$. I am thinking of a contour like this:
where $R=1$. If OP only requires the upper half, then we can disregard the $1$ in the final sum.
If you're taking $gamma$ to be this contour, note that $gamma$ is formed by two curves: $gamma_1$ and $gamma_2$ where $gamma_1$ is the interval $[-1,1]$ and $gamma_2$ is the upper arc of the semicircle. Thus we have that $$ intlimits_gamma |z| dz = intlimits_{gamma_1} |z|dz + intlimits_{gamma_2}|z|dz$$
You already found that $$ intlimits_{gamma_1} |z|dz = intlimits_{-1}^1 |z| dz = 1$$
We have to find $intlimits_{gamma_2}|z|dz$. Note that for all $z$ on $gamma_2$, $|z|=1$ because $z$ lies on the arc of the unit circle. So begin{align*}intlimits_{gamma_2}|z|dz = intlimits_{gamma_2}1dz = intlimits_0^pi ie^{itheta}dtheta = -2
end{align*}
So $$ intlimits_gamma |z| dz = -1$$
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
$endgroup$
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
add a comment |
$begingroup$
Note: I think I may have been considering a different contour, since the original poster asked what they could do after finding $intlimits_{-1}^1 |z|dz$. I am thinking of a contour like this:
where $R=1$. If OP only requires the upper half, then we can disregard the $1$ in the final sum.
If you're taking $gamma$ to be this contour, note that $gamma$ is formed by two curves: $gamma_1$ and $gamma_2$ where $gamma_1$ is the interval $[-1,1]$ and $gamma_2$ is the upper arc of the semicircle. Thus we have that $$ intlimits_gamma |z| dz = intlimits_{gamma_1} |z|dz + intlimits_{gamma_2}|z|dz$$
You already found that $$ intlimits_{gamma_1} |z|dz = intlimits_{-1}^1 |z| dz = 1$$
We have to find $intlimits_{gamma_2}|z|dz$. Note that for all $z$ on $gamma_2$, $|z|=1$ because $z$ lies on the arc of the unit circle. So begin{align*}intlimits_{gamma_2}|z|dz = intlimits_{gamma_2}1dz = intlimits_0^pi ie^{itheta}dtheta = -2
end{align*}
So $$ intlimits_gamma |z| dz = -1$$
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
$endgroup$
Note: I think I may have been considering a different contour, since the original poster asked what they could do after finding $intlimits_{-1}^1 |z|dz$. I am thinking of a contour like this:
where $R=1$. If OP only requires the upper half, then we can disregard the $1$ in the final sum.
If you're taking $gamma$ to be this contour, note that $gamma$ is formed by two curves: $gamma_1$ and $gamma_2$ where $gamma_1$ is the interval $[-1,1]$ and $gamma_2$ is the upper arc of the semicircle. Thus we have that $$ intlimits_gamma |z| dz = intlimits_{gamma_1} |z|dz + intlimits_{gamma_2}|z|dz$$
You already found that $$ intlimits_{gamma_1} |z|dz = intlimits_{-1}^1 |z| dz = 1$$
We have to find $intlimits_{gamma_2}|z|dz$. Note that for all $z$ on $gamma_2$, $|z|=1$ because $z$ lies on the arc of the unit circle. So begin{align*}intlimits_{gamma_2}|z|dz = intlimits_{gamma_2}1dz = intlimits_0^pi ie^{itheta}dtheta = -2
end{align*}
So $$ intlimits_gamma |z| dz = -1$$
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
edited Nov 28 '18 at 19:26
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
answered Nov 28 '18 at 19:06
J. PistachioJ. Pistachio
486212
486212
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
add a comment |
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
That's the contour I meant.
$endgroup$
– user3132457
Nov 28 '18 at 19:17
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
Apologies, I made a dumb error, let me fix that
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:23
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
$begingroup$
I just don't understand how you get the π. Isn't $int_{gamma}1dz=z$, which is just 1?
$endgroup$
– user3132457
Nov 29 '18 at 5:28
add a comment |
$begingroup$
Are you integrating clockwise or anti-clockwise?
Assuming you're integrating anti-clockwise, use a substitution of $z=mathrm e^{mathrm i theta}$, where $0 le theta le pi$.
It follows that $mathrm dz = mathrm{ie}^{mathrm i theta}~mathrm dtheta$.
Hence $$int_C |z|~mathrm dz = int_0^{pi} |mathrm e^{mathrm i theta}| ~mathrm{ie}^{mathrm i theta}~mathrm dtheta = mathrm iint_0^{pi}mathrm e^{mathrm itheta}~mathrm dtheta$$
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
$endgroup$
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
add a comment |
$begingroup$
Are you integrating clockwise or anti-clockwise?
Assuming you're integrating anti-clockwise, use a substitution of $z=mathrm e^{mathrm i theta}$, where $0 le theta le pi$.
It follows that $mathrm dz = mathrm{ie}^{mathrm i theta}~mathrm dtheta$.
Hence $$int_C |z|~mathrm dz = int_0^{pi} |mathrm e^{mathrm i theta}| ~mathrm{ie}^{mathrm i theta}~mathrm dtheta = mathrm iint_0^{pi}mathrm e^{mathrm itheta}~mathrm dtheta$$
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
$endgroup$
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
add a comment |
$begingroup$
Are you integrating clockwise or anti-clockwise?
Assuming you're integrating anti-clockwise, use a substitution of $z=mathrm e^{mathrm i theta}$, where $0 le theta le pi$.
It follows that $mathrm dz = mathrm{ie}^{mathrm i theta}~mathrm dtheta$.
Hence $$int_C |z|~mathrm dz = int_0^{pi} |mathrm e^{mathrm i theta}| ~mathrm{ie}^{mathrm i theta}~mathrm dtheta = mathrm iint_0^{pi}mathrm e^{mathrm itheta}~mathrm dtheta$$
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
$endgroup$
Are you integrating clockwise or anti-clockwise?
Assuming you're integrating anti-clockwise, use a substitution of $z=mathrm e^{mathrm i theta}$, where $0 le theta le pi$.
It follows that $mathrm dz = mathrm{ie}^{mathrm i theta}~mathrm dtheta$.
Hence $$int_C |z|~mathrm dz = int_0^{pi} |mathrm e^{mathrm i theta}| ~mathrm{ie}^{mathrm i theta}~mathrm dtheta = mathrm iint_0^{pi}mathrm e^{mathrm itheta}~mathrm dtheta$$
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
edited Nov 28 '18 at 19:16
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
answered Nov 28 '18 at 19:08
Fly by NightFly by Night
25.8k32978
25.8k32978
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
add a comment |
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
Given that the original integral is $int_{-1}^1$, I suspect the semicircle is actually supposed to be parametrized clockwise. (Also because your parametrization gives an answer of $-2$.)
$endgroup$
– André 3000
Nov 28 '18 at 19:11
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
$begingroup$
@André3000 I'm not sure. It's not clear from the OP. Most students, when integrating along the real line, have the lower limit less than the upper limit. (They don't realise that we're integrating along an oriented path.)
$endgroup$
– Fly by Night
Nov 28 '18 at 19:13
add a comment |
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$begingroup$
What is the upper semi-sphere? We're on a plane here.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:00
$begingroup$
Are you looking for $intlimits_gamma |z| dz$ where $gamma$ is semicircle in the upper half-plane?
$endgroup$
– J. Pistachio
Nov 28 '18 at 19:01
$begingroup$
Do you mean semi circle?
$endgroup$
– Cloud JR
Nov 28 '18 at 19:01
$begingroup$
Yes, sorry, semi circle. Will update the post now. @Josh, yes
$endgroup$
– user3132457
Nov 28 '18 at 19:02