Mixing
On the Poisson integral formula for the upper half plane
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I was trying to derive the Poisson formula for the upper half plane. However, I got stuck and found this Derive the Poisson Formula for a bounded C-harmonic function in the upper half-plane.
Now I am not able to prove the inequality $|(zeta - z)(zeta -overline{z})|>frac{1}{2}R^2$ which is required as per the answer by @mrf. I think its simple and straightforward but somehow I am missing out something. It will be really helpful if someone can provide me the details. Thanks in advance.
complex-analysis
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
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add a comment |
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I was trying to derive the Poisson formula for the upper half plane. However, I got stuck and found this Derive the Poisson Formula for a bounded C-harmonic function in the upper half-plane.
Now I am not able to prove the inequality $|(zeta - z)(zeta -overline{z})|>frac{1}{2}R^2$ which is required as per the answer by @mrf. I think its simple and straightforward but somehow I am missing out something. It will be really helpful if someone can provide me the details. Thanks in advance.
complex-analysis
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
$endgroup$
1
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
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– reuns
Jan 20 at 2:13
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38
add a comment |
$begingroup$
I was trying to derive the Poisson formula for the upper half plane. However, I got stuck and found this Derive the Poisson Formula for a bounded C-harmonic function in the upper half-plane.
Now I am not able to prove the inequality $|(zeta - z)(zeta -overline{z})|>frac{1}{2}R^2$ which is required as per the answer by @mrf. I think its simple and straightforward but somehow I am missing out something. It will be really helpful if someone can provide me the details. Thanks in advance.
complex-analysis
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
$endgroup$
I was trying to derive the Poisson formula for the upper half plane. However, I got stuck and found this Derive the Poisson Formula for a bounded C-harmonic function in the upper half-plane.
Now I am not able to prove the inequality $|(zeta - z)(zeta -overline{z})|>frac{1}{2}R^2$ which is required as per the answer by @mrf. I think its simple and straightforward but somehow I am missing out something. It will be really helpful if someone can provide me the details. Thanks in advance.
complex-analysis
complex-analysis
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
asked Jan 19 at 16:21
Himadri KarmakarHimadri Karmakar
1419
1419
1
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
$endgroup$
– reuns
Jan 20 at 2:13
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38
add a comment |
1
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
$endgroup$
– reuns
Jan 20 at 2:13
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38
1
1
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
$endgroup$
– reuns
Jan 20 at 2:13
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
$endgroup$
– reuns
Jan 20 at 2:13
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38
add a comment |
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1
$begingroup$
$z$ is fixed in the upper half-plane and $zeta$ is on the disk of radius $R$ so $|zeta-z|^2 ge (R-|z|)^2$, that's all you need.For $R$ large enough $(R-|z|)^2ge frac{R^2}{2} $
$endgroup$
– reuns
Jan 20 at 2:13
$begingroup$
@reuns Thanks. I got it.
$endgroup$
– Himadri Karmakar
Jan 20 at 3:38