Calculate the complex integral












0












$begingroup$


I have
$$
int{frac{dz}{z^2+9}}
$$
Also I'm given 2 different conditions.
First is $|z|=pi$, second is $|z-2i|=2$.



Okay, so for the integral i have $int{frac{dz}{(z+3i)(z-3i)}}$.



For the first condition, if I draw a circle, then $pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.



Please shed some light here. Thanks.






complex-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    The poles are at $pm 3i$. In each case, determine which poles are within the circle.
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 4:36










  • $begingroup$
    In both cases $3i$? If I draw that's what I get.
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:39
















0












$begingroup$


I have
$$
int{frac{dz}{z^2+9}}
$$
Also I'm given 2 different conditions.
First is $|z|=pi$, second is $|z-2i|=2$.



Okay, so for the integral i have $int{frac{dz}{(z+3i)(z-3i)}}$.



For the first condition, if I draw a circle, then $pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.



Please shed some light here. Thanks.






complex-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    The poles are at $pm 3i$. In each case, determine which poles are within the circle.
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 4:36










  • $begingroup$
    In both cases $3i$? If I draw that's what I get.
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:39














0












0








0





$begingroup$


I have
$$
int{frac{dz}{z^2+9}}
$$
Also I'm given 2 different conditions.
First is $|z|=pi$, second is $|z-2i|=2$.



Okay, so for the integral i have $int{frac{dz}{(z+3i)(z-3i)}}$.



For the first condition, if I draw a circle, then $pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.



Please shed some light here. Thanks.






complex-integration







share|cite|improve this question









$endgroup$




I have
$$
int{frac{dz}{z^2+9}}
$$
Also I'm given 2 different conditions.
First is $|z|=pi$, second is $|z-2i|=2$.



Okay, so for the integral i have $int{frac{dz}{(z+3i)(z-3i)}}$.



For the first condition, if I draw a circle, then $pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.



Please shed some light here. Thanks.






complex-integration




complex-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 4:33









user3132457user3132457

1598




1598












  • $begingroup$
    The poles are at $pm 3i$. In each case, determine which poles are within the circle.
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 4:36










  • $begingroup$
    In both cases $3i$? If I draw that's what I get.
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:39


















  • $begingroup$
    The poles are at $pm 3i$. In each case, determine which poles are within the circle.
    $endgroup$
    – Lord Shark the Unknown
    Dec 18 '18 at 4:36










  • $begingroup$
    In both cases $3i$? If I draw that's what I get.
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:39
















$begingroup$
The poles are at $pm 3i$. In each case, determine which poles are within the circle.
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:36




$begingroup$
The poles are at $pm 3i$. In each case, determine which poles are within the circle.
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:36












$begingroup$
In both cases $3i$? If I draw that's what I get.
$endgroup$
– user3132457
Dec 18 '18 at 4:39




$begingroup$
In both cases $3i$? If I draw that's what I get.
$endgroup$
– user3132457
Dec 18 '18 at 4:39










2 Answers
2






active

oldest

votes


















0












$begingroup$

For $mid zmid=pi$ the poles are both inside the circle. For $mid z-2imid =2$ only $3i$ is within the region.



Now use the residue theorem. That is, $f(z)=frac1{z^2+9}=frac1{6i}(frac1{z-3i}-frac1{z+3i})$. The residue at $3i$ is $frac1{6i}$. So the second integral is $2pi icdotfrac1{6i}=frac{pi}3$.



As for the first, we get the difference of the integrals over two smaller contours, both of which are $frac{pi}3$, so $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(z)={frac{1}{z^2+9}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:18










  • $begingroup$
    Right you are. Fixed it.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:21










  • $begingroup$
    Thanks, and how did you expand $f(z)$?
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:22










  • $begingroup$
    I factored the denominator. Then did sort of a partial fractions.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:26



















0












$begingroup$

For $|z|=pi$ the residue theorem implies that $int frac{1}{(z+3i)(z-3i)}dz = 2pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:48













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

For $mid zmid=pi$ the poles are both inside the circle. For $mid z-2imid =2$ only $3i$ is within the region.



Now use the residue theorem. That is, $f(z)=frac1{z^2+9}=frac1{6i}(frac1{z-3i}-frac1{z+3i})$. The residue at $3i$ is $frac1{6i}$. So the second integral is $2pi icdotfrac1{6i}=frac{pi}3$.



As for the first, we get the difference of the integrals over two smaller contours, both of which are $frac{pi}3$, so $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(z)={frac{1}{z^2+9}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:18










  • $begingroup$
    Right you are. Fixed it.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:21










  • $begingroup$
    Thanks, and how did you expand $f(z)$?
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:22










  • $begingroup$
    I factored the denominator. Then did sort of a partial fractions.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:26
















0












$begingroup$

For $mid zmid=pi$ the poles are both inside the circle. For $mid z-2imid =2$ only $3i$ is within the region.



Now use the residue theorem. That is, $f(z)=frac1{z^2+9}=frac1{6i}(frac1{z-3i}-frac1{z+3i})$. The residue at $3i$ is $frac1{6i}$. So the second integral is $2pi icdotfrac1{6i}=frac{pi}3$.



As for the first, we get the difference of the integrals over two smaller contours, both of which are $frac{pi}3$, so $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(z)={frac{1}{z^2+9}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:18










  • $begingroup$
    Right you are. Fixed it.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:21










  • $begingroup$
    Thanks, and how did you expand $f(z)$?
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:22










  • $begingroup$
    I factored the denominator. Then did sort of a partial fractions.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:26














0












0








0





$begingroup$

For $mid zmid=pi$ the poles are both inside the circle. For $mid z-2imid =2$ only $3i$ is within the region.



Now use the residue theorem. That is, $f(z)=frac1{z^2+9}=frac1{6i}(frac1{z-3i}-frac1{z+3i})$. The residue at $3i$ is $frac1{6i}$. So the second integral is $2pi icdotfrac1{6i}=frac{pi}3$.



As for the first, we get the difference of the integrals over two smaller contours, both of which are $frac{pi}3$, so $0$.






share|cite|improve this answer











$endgroup$



For $mid zmid=pi$ the poles are both inside the circle. For $mid z-2imid =2$ only $3i$ is within the region.



Now use the residue theorem. That is, $f(z)=frac1{z^2+9}=frac1{6i}(frac1{z-3i}-frac1{z+3i})$. The residue at $3i$ is $frac1{6i}$. So the second integral is $2pi icdotfrac1{6i}=frac{pi}3$.



As for the first, we get the difference of the integrals over two smaller contours, both of which are $frac{pi}3$, so $0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 5:19

























answered Dec 18 '18 at 5:14









Chris CusterChris Custer

13.7k3827




13.7k3827












  • $begingroup$
    $f(z)={frac{1}{z^2+9}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:18










  • $begingroup$
    Right you are. Fixed it.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:21










  • $begingroup$
    Thanks, and how did you expand $f(z)$?
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:22










  • $begingroup$
    I factored the denominator. Then did sort of a partial fractions.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:26


















  • $begingroup$
    $f(z)={frac{1}{z^2+9}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:18










  • $begingroup$
    Right you are. Fixed it.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:21










  • $begingroup$
    Thanks, and how did you expand $f(z)$?
    $endgroup$
    – user3132457
    Dec 18 '18 at 5:22










  • $begingroup$
    I factored the denominator. Then did sort of a partial fractions.
    $endgroup$
    – Chris Custer
    Dec 18 '18 at 5:26
















$begingroup$
$f(z)={frac{1}{z^2+9}}$
$endgroup$
– user3132457
Dec 18 '18 at 5:18




$begingroup$
$f(z)={frac{1}{z^2+9}}$
$endgroup$
– user3132457
Dec 18 '18 at 5:18












$begingroup$
Right you are. Fixed it.
$endgroup$
– Chris Custer
Dec 18 '18 at 5:21




$begingroup$
Right you are. Fixed it.
$endgroup$
– Chris Custer
Dec 18 '18 at 5:21












$begingroup$
Thanks, and how did you expand $f(z)$?
$endgroup$
– user3132457
Dec 18 '18 at 5:22




$begingroup$
Thanks, and how did you expand $f(z)$?
$endgroup$
– user3132457
Dec 18 '18 at 5:22












$begingroup$
I factored the denominator. Then did sort of a partial fractions.
$endgroup$
– Chris Custer
Dec 18 '18 at 5:26




$begingroup$
I factored the denominator. Then did sort of a partial fractions.
$endgroup$
– Chris Custer
Dec 18 '18 at 5:26











0












$begingroup$

For $|z|=pi$ the residue theorem implies that $int frac{1}{(z+3i)(z-3i)}dz = 2pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:48


















0












$begingroup$

For $|z|=pi$ the residue theorem implies that $int frac{1}{(z+3i)(z-3i)}dz = 2pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:48
















0












0








0





$begingroup$

For $|z|=pi$ the residue theorem implies that $int frac{1}{(z+3i)(z-3i)}dz = 2pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.






share|cite|improve this answer









$endgroup$



For $|z|=pi$ the residue theorem implies that $int frac{1}{(z+3i)(z-3i)}dz = 2pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 4:40









Mustafa SaidMustafa Said

3,0011913




3,0011913












  • $begingroup$
    This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:48




















  • $begingroup$
    This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
    $endgroup$
    – user3132457
    Dec 18 '18 at 4:48


















$begingroup$
This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
$endgroup$
– user3132457
Dec 18 '18 at 4:48






$begingroup$
This is my solution for 2nd one but the answer is incorrect. $$ int{frac{dz}{z^2+9}} = int{frac{dz}{(z+3i)(z-3i)}} = int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{2pi i }} int {frac{{frac{dz}{z+3i}}}{z-3i}} = 2pi i {frac{1}{z+3i}} prime (at 3i) = {frac{-2pi i times (-1)}{(z+3i)^2}} = {frac{-2pi i}{(3i+3i)^2}} = {frac{-2pi i}{9+18i-9}} = {frac{-2pi}{-18i}} = {frac{pi}{9}} $$ The answer is ${frac{pi}{3}}$
$endgroup$
– user3132457
Dec 18 '18 at 4:48




















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