How do I perform this complex integration?
up vote
2
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Evaluate the complex integration $(z^2 + 3z)$ wrt $z$ along the circle $|z| = 2$, from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
As far as I understand, this can be solved by taking $x = 2 cos theta$, $y = 2 sin theta$, and then integrating wrt $theta$ from $0$ to $π/2$.
But on making these substitutions, the integration becomes quite lengthy and clumsy.
Is there any other way to solve this, which I might be missing right now?
complex-systems
edited Nov 17 at 3:27
Tianlalu
2,8811935
asked Nov 17 at 3:22
Jasmine
111
add a comment |
up vote
2
down vote
favorite
Evaluate the complex integration $(z^2 + 3z)$ wrt $z$ along the circle $|z| = 2$, from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
As far as I understand, this can be solved by taking $x = 2 cos theta$, $y = 2 sin theta$, and then integrating wrt $theta$ from $0$ to $π/2$.
But on making these substitutions, the integration becomes quite lengthy and clumsy.
Is there any other way to solve this, which I might be missing right now?
complex-systems
edited Nov 17 at 3:27
Tianlalu
2,8811935
asked Nov 17 at 3:22
Jasmine
111
Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Evaluate the complex integration $(z^2 + 3z)$ wrt $z$ along the circle $|z| = 2$, from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
As far as I understand, this can be solved by taking $x = 2 cos theta$, $y = 2 sin theta$, and then integrating wrt $theta$ from $0$ to $π/2$.
But on making these substitutions, the integration becomes quite lengthy and clumsy.
Is there any other way to solve this, which I might be missing right now?
complex-systems
edited Nov 17 at 3:27
Tianlalu
2,8811935
asked Nov 17 at 3:22
Jasmine
111
Evaluate the complex integration $(z^2 + 3z)$ wrt $z$ along the circle $|z| = 2$, from $(2,0)$ to $(0,2)$ in a counterclockwise direction.
As far as I understand, this can be solved by taking $x = 2 cos theta$, $y = 2 sin theta$, and then integrating wrt $theta$ from $0$ to $π/2$.
But on making these substitutions, the integration becomes quite lengthy and clumsy.
Is there any other way to solve this, which I might be missing right now?
complex-systems
complex-systems
edited Nov 17 at 3:27
Tianlalu
2,8811935
asked Nov 17 at 3:22
Jasmine
111
edited Nov 17 at 3:27
Tianlalu
2,8811935
asked Nov 17 at 3:22
Jasmine
111
edited Nov 17 at 3:27
Tianlalu
2,8811935
edited Nov 17 at 3:27
Tianlalu
2,8811935
edited Nov 17 at 3:27
Tianlalu
2,8811935
2,8811935
asked Nov 17 at 3:22
Jasmine
111
asked Nov 17 at 3:22
Jasmine
111
asked Nov 17 at 3:22
Jasmine
111
111
Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33
add a comment |
Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33
Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33
Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33
add a comment |
2 Answers
2
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oldest
votes
up vote
2
down vote
$$ int _{2}^{2i} z^2+3z dz =(1/3)z^3 +(3/2) z^2 |_2^{2i} = (-44/3)-(8/3)i$$
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
|
show 2 more comments
up vote
1
down vote
On your circle, $z=2e^{it}$, $0leq tleqpi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it},dt$)
begin{align}
int_0^{pi/2} ((2e^{it})^2+6e^{it}),2ie^{it},dt
&=2iint_0^{pi/2}(4e^{3it}+6e^{2it}),dt
=2ileft[ left.frac{4e^{3it}}{3i}+frac{3e^{2it}}{i}right|_0^{pi/2}right]\ \
&=-frac {8i}3-6-frac83-6=-frac {44}3-frac {8i}3.
end{align}
answered Nov 17 at 3:33
Martin Argerami
122k1174172
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$ int _{2}^{2i} z^2+3z dz =(1/3)z^3 +(3/2) z^2 |_2^{2i} = (-44/3)-(8/3)i$$
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
|
show 2 more comments
up vote
2
down vote
$$ int _{2}^{2i} z^2+3z dz =(1/3)z^3 +(3/2) z^2 |_2^{2i} = (-44/3)-(8/3)i$$
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
|
show 2 more comments
up vote
2
down vote
up vote
2
down vote
$$ int _{2}^{2i} z^2+3z dz =(1/3)z^3 +(3/2) z^2 |_2^{2i} = (-44/3)-(8/3)i$$
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
$$ int _{2}^{2i} z^2+3z dz =(1/3)z^3 +(3/2) z^2 |_2^{2i} = (-44/3)-(8/3)i$$
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
edited Nov 17 at 5:56
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 4:01
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
|
show 2 more comments
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
This is easy ! I had thought of this earlier, but what confused me was the circle given in question.
– Jasmine
Nov 17 at 4:18
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@Jasmine: This is because of "The Fundamental Theorem of Calculus".
– Yadati Kiran
Nov 17 at 4:24
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@epimorphic: I do agree that the Fundamental Theorem of Calculus cannot be applied to complex contour integrals but its the most efficient way to get the result provided the domain is simply connected and the function is analytic on the domain
– Yadati Kiran
Nov 17 at 5:37
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@Mohammad Riazi-Kermani: Your integrand should be $z^2+3z$.
– quasi
Nov 17 at 5:49
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
@YadatiKiran Oops, you're right, I really need to review my complex analysis... *slinks back into corner*
– epimorphic
Nov 17 at 5:50
|
show 2 more comments
up vote
1
down vote
On your circle, $z=2e^{it}$, $0leq tleqpi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it},dt$)
begin{align}
int_0^{pi/2} ((2e^{it})^2+6e^{it}),2ie^{it},dt
&=2iint_0^{pi/2}(4e^{3it}+6e^{2it}),dt
=2ileft[ left.frac{4e^{3it}}{3i}+frac{3e^{2it}}{i}right|_0^{pi/2}right]\ \
&=-frac {8i}3-6-frac83-6=-frac {44}3-frac {8i}3.
end{align}
answered Nov 17 at 3:33
Martin Argerami
122k1174172
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
|
show 1 more comment
up vote
1
down vote
On your circle, $z=2e^{it}$, $0leq tleqpi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it},dt$)
begin{align}
int_0^{pi/2} ((2e^{it})^2+6e^{it}),2ie^{it},dt
&=2iint_0^{pi/2}(4e^{3it}+6e^{2it}),dt
=2ileft[ left.frac{4e^{3it}}{3i}+frac{3e^{2it}}{i}right|_0^{pi/2}right]\ \
&=-frac {8i}3-6-frac83-6=-frac {44}3-frac {8i}3.
end{align}
answered Nov 17 at 3:33
Martin Argerami
122k1174172
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
|
show 1 more comment
up vote
1
down vote
up vote
1
down vote
On your circle, $z=2e^{it}$, $0leq tleqpi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it},dt$)
begin{align}
int_0^{pi/2} ((2e^{it})^2+6e^{it}),2ie^{it},dt
&=2iint_0^{pi/2}(4e^{3it}+6e^{2it}),dt
=2ileft[ left.frac{4e^{3it}}{3i}+frac{3e^{2it}}{i}right|_0^{pi/2}right]\ \
&=-frac {8i}3-6-frac83-6=-frac {44}3-frac {8i}3.
end{align}
answered Nov 17 at 3:33
Martin Argerami
122k1174172
On your circle, $z=2e^{it}$, $0leq tleqpi/2$. Then your integral is (never forget to include the change of variable, $dz=2ie^{it},dt$)
begin{align}
int_0^{pi/2} ((2e^{it})^2+6e^{it}),2ie^{it},dt
&=2iint_0^{pi/2}(4e^{3it}+6e^{2it}),dt
=2ileft[ left.frac{4e^{3it}}{3i}+frac{3e^{2it}}{i}right|_0^{pi/2}right]\ \
&=-frac {8i}3-6-frac83-6=-frac {44}3-frac {8i}3.
end{align}
answered Nov 17 at 3:33
Martin Argerami
122k1174172
edited Nov 17 at 6:08
answered Nov 17 at 3:33
Martin Argerami
122k1174172
answered Nov 17 at 3:33
Martin Argerami
122k1174172
answered Nov 17 at 3:33
Martin Argerami
122k1174172
122k1174172
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
|
show 1 more comment
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
2
2
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
Yes. Now I get it. Instead of writing z as x + iy, I should write it in the r e^it form, which makes the task simpler.
– Jasmine
Nov 17 at 3:46
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
@MartinArgerami You lost a factor of $e^{it}$ going between the two integral expressions.
– epimorphic
Nov 17 at 5:04
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
Indeed, thanks for noticing!
– Martin Argerami
Nov 17 at 5:41
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
+1, but still some errors. The final answer should be $$-frac{44}{3}-frac{8}{3}i$$
– quasi
Nov 17 at 5:47
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
Indeed. Thanks.
– Martin Argerami
Nov 17 at 6:08
|
show 1 more comment
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Try a contour integral, using (0,0) to (2,0) and (0,2) to (0,0) as additional sides. and note that the contour integral has no poles.
– Andrei
Nov 17 at 3:33