is ln r or ln|z| harmonic in C [closed]
$begingroup$
ln|z| is harmonic in C because it satisfy laplace equation n also partial derivatives exist n continuous but here on book it is mentioned that ln|z| is harmonic in punctured disk C-{0}...how? is it because log function is defined for positive value of x ie x>0
next ln|z| has no harmonic conjugate in C-{0} though it does have in C-[0,infinity) ..how ..we have milne thomson method to find its harmonic conjugate v but in this case m not being able to calculate
complex-analysis harmonic-functions analytic-functions
asked Jan 8 at 14:51
HenryHenry
357
$endgroup$
closed as unclear what you're asking by Xander Henderson, Cesareo, Adrian Keister, Alexander Gruber♦ Jan 8 at 22:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
ln|z| is harmonic in C because it satisfy laplace equation n also partial derivatives exist n continuous but here on book it is mentioned that ln|z| is harmonic in punctured disk C-{0}...how? is it because log function is defined for positive value of x ie x>0
next ln|z| has no harmonic conjugate in C-{0} though it does have in C-[0,infinity) ..how ..we have milne thomson method to find its harmonic conjugate v but in this case m not being able to calculate
complex-analysis harmonic-functions analytic-functions
asked Jan 8 at 14:51
HenryHenry
357
$endgroup$
closed as unclear what you're asking by Xander Henderson, Cesareo, Adrian Keister, Alexander Gruber♦ Jan 8 at 22:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57
add a comment |
$begingroup$
ln|z| is harmonic in C because it satisfy laplace equation n also partial derivatives exist n continuous but here on book it is mentioned that ln|z| is harmonic in punctured disk C-{0}...how? is it because log function is defined for positive value of x ie x>0
next ln|z| has no harmonic conjugate in C-{0} though it does have in C-[0,infinity) ..how ..we have milne thomson method to find its harmonic conjugate v but in this case m not being able to calculate
complex-analysis harmonic-functions analytic-functions
asked Jan 8 at 14:51
HenryHenry
357
$endgroup$
ln|z| is harmonic in C because it satisfy laplace equation n also partial derivatives exist n continuous but here on book it is mentioned that ln|z| is harmonic in punctured disk C-{0}...how? is it because log function is defined for positive value of x ie x>0
next ln|z| has no harmonic conjugate in C-{0} though it does have in C-[0,infinity) ..how ..we have milne thomson method to find its harmonic conjugate v but in this case m not being able to calculate
complex-analysis harmonic-functions analytic-functions
complex-analysis harmonic-functions analytic-functions
asked Jan 8 at 14:51
HenryHenry
357
asked Jan 8 at 14:51
HenryHenry
357
asked Jan 8 at 14:51
HenryHenry
357
asked Jan 8 at 14:51
HenryHenry
357
asked Jan 8 at 14:51
HenryHenry
357
357
closed as unclear what you're asking by Xander Henderson, Cesareo, Adrian Keister, Alexander Gruber♦ Jan 8 at 22:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Xander Henderson, Cesareo, Adrian Keister, Alexander Gruber♦ Jan 8 at 22:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57
add a comment |
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57
add a comment |
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $log$ function is not defined at $|z|=0$.
$endgroup$
– Jerry
Jan 8 at 14:56
$begingroup$
If a function is harmonic on an open set then it is harmonic on any open subset. So if you claim that it is harmonic on $mathbb C$ (which it isn't) you shouldn't be confused about how its harmonic on $mathbb Csetminus {0}$(which isn't a disk?)
$endgroup$
– Calvin Khor
Jan 8 at 14:57