Find the complex integral












0












$begingroup$


I have
$$int_{gamma} {frac{z}{overline z}}dz$$
where $gamma$ is the edge of ${1 < |z| < 2 $and $Im (z) > 0}$.



I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.



For $|z|=1$, the integral is $ipi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-ipi$. But the answer in book is ${frac{4}{3}}$
.






complex-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
    $endgroup$
    – CyclotomicField
    Nov 29 '18 at 18:51
















0












$begingroup$


I have
$$int_{gamma} {frac{z}{overline z}}dz$$
where $gamma$ is the edge of ${1 < |z| < 2 $and $Im (z) > 0}$.



I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.



For $|z|=1$, the integral is $ipi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-ipi$. But the answer in book is ${frac{4}{3}}$
.






complex-integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
    $endgroup$
    – CyclotomicField
    Nov 29 '18 at 18:51














0












0








0





$begingroup$


I have
$$int_{gamma} {frac{z}{overline z}}dz$$
where $gamma$ is the edge of ${1 < |z| < 2 $and $Im (z) > 0}$.



I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.



For $|z|=1$, the integral is $ipi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-ipi$. But the answer in book is ${frac{4}{3}}$
.






complex-integration







share|cite|improve this question









$endgroup$




I have
$$int_{gamma} {frac{z}{overline z}}dz$$
where $gamma$ is the edge of ${1 < |z| < 2 $and $Im (z) > 0}$.



I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.



For $|z|=1$, the integral is $ipi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-ipi$. But the answer in book is ${frac{4}{3}}$
.






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 Nov 29 '18 at 18:46









user3132457user3132457

1548




1548












  • $begingroup$
    Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
    $endgroup$
    – CyclotomicField
    Nov 29 '18 at 18:51


















  • $begingroup$
    Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
    $endgroup$
    – CyclotomicField
    Nov 29 '18 at 18:51
















$begingroup$
Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
$endgroup$
– CyclotomicField
Nov 29 '18 at 18:51




$begingroup$
Note that $frac{z}{bar{z}} =frac{z^2}{|z|^2}$.
$endgroup$
– CyclotomicField
Nov 29 '18 at 18:51










1 Answer
1






active

oldest

votes


















0












$begingroup$

Actually, in the upper semi-circle $lvert zrvert=1$, you have$$frac z{overline z}=frac{z^2}{lvert zrvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-dfrac43$.



Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:06












  • $begingroup$
    I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:08










  • $begingroup$
    For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:19












  • $begingroup$
    @user3132457 Yes, I made a mistake. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:22










  • $begingroup$
    Is it -4/3 or 4/3?
    $endgroup$
    – user3132457
    Nov 30 '18 at 5:18











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Actually, in the upper semi-circle $lvert zrvert=1$, you have$$frac z{overline z}=frac{z^2}{lvert zrvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-dfrac43$.



Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:06












  • $begingroup$
    I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:08










  • $begingroup$
    For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:19












  • $begingroup$
    @user3132457 Yes, I made a mistake. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:22










  • $begingroup$
    Is it -4/3 or 4/3?
    $endgroup$
    – user3132457
    Nov 30 '18 at 5:18
















0












$begingroup$

Actually, in the upper semi-circle $lvert zrvert=1$, you have$$frac z{overline z}=frac{z^2}{lvert zrvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-dfrac43$.



Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:06












  • $begingroup$
    I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:08










  • $begingroup$
    For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:19












  • $begingroup$
    @user3132457 Yes, I made a mistake. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:22










  • $begingroup$
    Is it -4/3 or 4/3?
    $endgroup$
    – user3132457
    Nov 30 '18 at 5:18














0












0








0





$begingroup$

Actually, in the upper semi-circle $lvert zrvert=1$, you have$$frac z{overline z}=frac{z^2}{lvert zrvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-dfrac43$.



Can you take it from here?






share|cite|improve this answer











$endgroup$



Actually, in the upper semi-circle $lvert zrvert=1$, you have$$frac z{overline z}=frac{z^2}{lvert zrvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-dfrac43$.



Can you take it from here?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 19:21

























answered Nov 29 '18 at 18:55









José Carlos SantosJosé Carlos Santos

159k22126231




159k22126231












  • $begingroup$
    I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:06












  • $begingroup$
    I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:08










  • $begingroup$
    For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:19












  • $begingroup$
    @user3132457 Yes, I made a mistake. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:22










  • $begingroup$
    Is it -4/3 or 4/3?
    $endgroup$
    – user3132457
    Nov 30 '18 at 5:18


















  • $begingroup$
    I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:06












  • $begingroup$
    I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:08










  • $begingroup$
    For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
    $endgroup$
    – user3132457
    Nov 29 '18 at 19:19












  • $begingroup$
    @user3132457 Yes, I made a mistake. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 19:22










  • $begingroup$
    Is it -4/3 or 4/3?
    $endgroup$
    – user3132457
    Nov 30 '18 at 5:18
















$begingroup$
I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
$endgroup$
– user3132457
Nov 29 '18 at 19:06






$begingroup$
I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? ibb.co/RYY7qBp
$endgroup$
– user3132457
Nov 29 '18 at 19:06














$begingroup$
I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 19:08




$begingroup$
I used the fact that a primitive of $z^2$ is $frac{z^3}3$. So, the first of those integrals is $frac{1^3}3-frac{(-1)^3}3=frac23$. The other one is similar.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 19:08












$begingroup$
For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
$endgroup$
– user3132457
Nov 29 '18 at 19:19






$begingroup$
For |z|=2, I get $$int {frac{z}{overline z}}dz = int {frac{z^2}{|z|^2}} = int {frac{z^2}{4}} = {frac{z^3}{3}} times {frac{1}{4}} = {frac{1}{12}}(8+8) = {frac{16}{12}} = {frac{4}{3}}$$ which looks like the final answer.
$endgroup$
– user3132457
Nov 29 '18 at 19:19














$begingroup$
@user3132457 Yes, I made a mistake. I've edited my answer.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 19:22




$begingroup$
@user3132457 Yes, I made a mistake. I've edited my answer.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 19:22












$begingroup$
Is it -4/3 or 4/3?
$endgroup$
– user3132457
Nov 30 '18 at 5:18




$begingroup$
Is it -4/3 or 4/3?
$endgroup$
– user3132457
Nov 30 '18 at 5:18


















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