Integral of a complex function over contour (continued)
$begingroup$
I'd like to double check that my method to evaluating the following integral over a contour is correct. I asked how to go about integrating over such a contour in Integral of a complex function over contour. The help I recieved was fantastic, though I still have a few questions regarding a general approach in evaluating such integrals.
$Gamma$ is a simple closed contour given by the path moving from 0 to 1 along the real axis, then from 1 to i along the unit circle, then from i to 0 along the imaginary axis.
We paramatrize $Gamma$ as follows:
- z(t)=t, from $0 leq t leq 1$ [0 to 1 via the real axis]
- z(t)=exp(it), from $0 leq t leq frac{pi}{2}$ [1 to i via the unit circle]
- z(t)=it, from $0 leq t leq 1$ [i to 0 via the imaginary axis]
From this we find that the integral
$$int_Gamma zdz$$ is 0, using the formula $int_{a}^{b}z(t)z'(t)dt$ and summing for the three paths written above.
My question is whether or not the following approach is correct for evaluating the integral:
$int_Gamma bar{z}dz$.
For the first path of the contour, I calculate $int_{0}^{1}bar{z}(t)bar{z}'(t)dt$. As mentioned above, in this case, $z(t)=t$ therefore $bar{z}(t)=t$, $bar{z}'(t)=1$ and $dz=dt$. So $int_{0}^{1}bar{z}(t)bar{z}'(t)dt=frac{1}{2}$.
For the second path, I calculate $int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt$ where $z(t)=exp(it)$ and so $bar{z}(t)=exp(-it)$, $bar{z}'(t)=-iexp(-it)$ and $dz=iexp(it)dt$. So $$int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt = int_{0}^{frac{pi}{2}}exp(-it)(-iexp(-it))(iexp(it))dt=int_{0}^{frac{pi}{2}}exp(-it)dt=1-i$$
For the third path I calculate $-int_{0}^{1}bar{z}(t)bar{z}'(t)dt$ (minus because of the orientation of the path) where $z(t)=it$ and so $bar{z}(t)=-it$, $bar{z}'(t)=-i$ and $dz=idt$. So $$-int_{0}^{1}bar{z}(t)bar{z}'(t)dt =-int_{0}^{1}(-it)(-i)i dt = iint_{0}^{1}tdt = frac{i}{2}$$
By summing all of these we get $frac{1}{2}+1-i+frac{i}{2}=frac{3-i}{2}$.
Is this correct, or at least is the method correct?
complex-analysis contour-integration complex-integration
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
$endgroup$
add a comment |
$begingroup$
I'd like to double check that my method to evaluating the following integral over a contour is correct. I asked how to go about integrating over such a contour in Integral of a complex function over contour. The help I recieved was fantastic, though I still have a few questions regarding a general approach in evaluating such integrals.
$Gamma$ is a simple closed contour given by the path moving from 0 to 1 along the real axis, then from 1 to i along the unit circle, then from i to 0 along the imaginary axis.
We paramatrize $Gamma$ as follows:
- z(t)=t, from $0 leq t leq 1$ [0 to 1 via the real axis]
- z(t)=exp(it), from $0 leq t leq frac{pi}{2}$ [1 to i via the unit circle]
- z(t)=it, from $0 leq t leq 1$ [i to 0 via the imaginary axis]
From this we find that the integral
$$int_Gamma zdz$$ is 0, using the formula $int_{a}^{b}z(t)z'(t)dt$ and summing for the three paths written above.
My question is whether or not the following approach is correct for evaluating the integral:
$int_Gamma bar{z}dz$.
For the first path of the contour, I calculate $int_{0}^{1}bar{z}(t)bar{z}'(t)dt$. As mentioned above, in this case, $z(t)=t$ therefore $bar{z}(t)=t$, $bar{z}'(t)=1$ and $dz=dt$. So $int_{0}^{1}bar{z}(t)bar{z}'(t)dt=frac{1}{2}$.
For the second path, I calculate $int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt$ where $z(t)=exp(it)$ and so $bar{z}(t)=exp(-it)$, $bar{z}'(t)=-iexp(-it)$ and $dz=iexp(it)dt$. So $$int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt = int_{0}^{frac{pi}{2}}exp(-it)(-iexp(-it))(iexp(it))dt=int_{0}^{frac{pi}{2}}exp(-it)dt=1-i$$
For the third path I calculate $-int_{0}^{1}bar{z}(t)bar{z}'(t)dt$ (minus because of the orientation of the path) where $z(t)=it$ and so $bar{z}(t)=-it$, $bar{z}'(t)=-i$ and $dz=idt$. So $$-int_{0}^{1}bar{z}(t)bar{z}'(t)dt =-int_{0}^{1}(-it)(-i)i dt = iint_{0}^{1}tdt = frac{i}{2}$$
By summing all of these we get $frac{1}{2}+1-i+frac{i}{2}=frac{3-i}{2}$.
Is this correct, or at least is the method correct?
complex-analysis contour-integration complex-integration
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
$endgroup$
1
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58
add a comment |
$begingroup$
I'd like to double check that my method to evaluating the following integral over a contour is correct. I asked how to go about integrating over such a contour in Integral of a complex function over contour. The help I recieved was fantastic, though I still have a few questions regarding a general approach in evaluating such integrals.
$Gamma$ is a simple closed contour given by the path moving from 0 to 1 along the real axis, then from 1 to i along the unit circle, then from i to 0 along the imaginary axis.
We paramatrize $Gamma$ as follows:
- z(t)=t, from $0 leq t leq 1$ [0 to 1 via the real axis]
- z(t)=exp(it), from $0 leq t leq frac{pi}{2}$ [1 to i via the unit circle]
- z(t)=it, from $0 leq t leq 1$ [i to 0 via the imaginary axis]
From this we find that the integral
$$int_Gamma zdz$$ is 0, using the formula $int_{a}^{b}z(t)z'(t)dt$ and summing for the three paths written above.
My question is whether or not the following approach is correct for evaluating the integral:
$int_Gamma bar{z}dz$.
For the first path of the contour, I calculate $int_{0}^{1}bar{z}(t)bar{z}'(t)dt$. As mentioned above, in this case, $z(t)=t$ therefore $bar{z}(t)=t$, $bar{z}'(t)=1$ and $dz=dt$. So $int_{0}^{1}bar{z}(t)bar{z}'(t)dt=frac{1}{2}$.
For the second path, I calculate $int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt$ where $z(t)=exp(it)$ and so $bar{z}(t)=exp(-it)$, $bar{z}'(t)=-iexp(-it)$ and $dz=iexp(it)dt$. So $$int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt = int_{0}^{frac{pi}{2}}exp(-it)(-iexp(-it))(iexp(it))dt=int_{0}^{frac{pi}{2}}exp(-it)dt=1-i$$
For the third path I calculate $-int_{0}^{1}bar{z}(t)bar{z}'(t)dt$ (minus because of the orientation of the path) where $z(t)=it$ and so $bar{z}(t)=-it$, $bar{z}'(t)=-i$ and $dz=idt$. So $$-int_{0}^{1}bar{z}(t)bar{z}'(t)dt =-int_{0}^{1}(-it)(-i)i dt = iint_{0}^{1}tdt = frac{i}{2}$$
By summing all of these we get $frac{1}{2}+1-i+frac{i}{2}=frac{3-i}{2}$.
Is this correct, or at least is the method correct?
complex-analysis contour-integration complex-integration
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
$endgroup$
I'd like to double check that my method to evaluating the following integral over a contour is correct. I asked how to go about integrating over such a contour in Integral of a complex function over contour. The help I recieved was fantastic, though I still have a few questions regarding a general approach in evaluating such integrals.
$Gamma$ is a simple closed contour given by the path moving from 0 to 1 along the real axis, then from 1 to i along the unit circle, then from i to 0 along the imaginary axis.
We paramatrize $Gamma$ as follows:
- z(t)=t, from $0 leq t leq 1$ [0 to 1 via the real axis]
- z(t)=exp(it), from $0 leq t leq frac{pi}{2}$ [1 to i via the unit circle]
- z(t)=it, from $0 leq t leq 1$ [i to 0 via the imaginary axis]
From this we find that the integral
$$int_Gamma zdz$$ is 0, using the formula $int_{a}^{b}z(t)z'(t)dt$ and summing for the three paths written above.
My question is whether or not the following approach is correct for evaluating the integral:
$int_Gamma bar{z}dz$.
For the first path of the contour, I calculate $int_{0}^{1}bar{z}(t)bar{z}'(t)dt$. As mentioned above, in this case, $z(t)=t$ therefore $bar{z}(t)=t$, $bar{z}'(t)=1$ and $dz=dt$. So $int_{0}^{1}bar{z}(t)bar{z}'(t)dt=frac{1}{2}$.
For the second path, I calculate $int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt$ where $z(t)=exp(it)$ and so $bar{z}(t)=exp(-it)$, $bar{z}'(t)=-iexp(-it)$ and $dz=iexp(it)dt$. So $$int_{0}^{frac{pi}{2}}bar{z}(t)bar{z}'(t)dt = int_{0}^{frac{pi}{2}}exp(-it)(-iexp(-it))(iexp(it))dt=int_{0}^{frac{pi}{2}}exp(-it)dt=1-i$$
For the third path I calculate $-int_{0}^{1}bar{z}(t)bar{z}'(t)dt$ (minus because of the orientation of the path) where $z(t)=it$ and so $bar{z}(t)=-it$, $bar{z}'(t)=-i$ and $dz=idt$. So $$-int_{0}^{1}bar{z}(t)bar{z}'(t)dt =-int_{0}^{1}(-it)(-i)i dt = iint_{0}^{1}tdt = frac{i}{2}$$
By summing all of these we get $frac{1}{2}+1-i+frac{i}{2}=frac{3-i}{2}$.
Is this correct, or at least is the method correct?
complex-analysis contour-integration complex-integration
complex-analysis contour-integration complex-integration
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
asked Nov 29 '18 at 13:20
basic_ceremonybasic_ceremony
53
53
1
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58
add a comment |
1
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58
1
1
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58
add a comment |
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1
$begingroup$
If your goal is to integrate $z$, why do you switch to $overline z$?
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:40
$begingroup$
The goal here is to integrate $bar{z}$ in this case. I was using the method of integrating z as a basis for solving the integral of $bar{z}$ and I wonder if this approach is correct.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:50
$begingroup$
I did not check the computations, but, yes, the method is correct.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 13:56
$begingroup$
Thank you very much, José. I appreciate the help.
$endgroup$
– basic_ceremony
Nov 29 '18 at 13:58