Show that $ |a_n| leq 2(1-alpha)$ for every $n$
$begingroup$
Suppose that $g(z)=1+sum_{n=1}^{infty}a_nz^n$ is analytic for $|z|<1$, with $Re(g(z))> alpha$. Show that $ |a_n| leq 2(1-alpha)$ for every $n$.
Answer:
Since $Re(g(z))>alpha$ implies $g(z)$ is bounded, say $|g(z)|>alpha$.
Then,
$g(z)=1+sum_{n=1}^{infty}a_nz^n \ Rightarrow |g(z)| leq 1+ sum |a_nz^n| $
any hints for showing this.
complex-analysis
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
$endgroup$
add a comment |
$begingroup$
Suppose that $g(z)=1+sum_{n=1}^{infty}a_nz^n$ is analytic for $|z|<1$, with $Re(g(z))> alpha$. Show that $ |a_n| leq 2(1-alpha)$ for every $n$.
Answer:
Since $Re(g(z))>alpha$ implies $g(z)$ is bounded, say $|g(z)|>alpha$.
Then,
$g(z)=1+sum_{n=1}^{infty}a_nz^n \ Rightarrow |g(z)| leq 1+ sum |a_nz^n| $
any hints for showing this.
complex-analysis
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
$endgroup$
add a comment |
$begingroup$
Suppose that $g(z)=1+sum_{n=1}^{infty}a_nz^n$ is analytic for $|z|<1$, with $Re(g(z))> alpha$. Show that $ |a_n| leq 2(1-alpha)$ for every $n$.
Answer:
Since $Re(g(z))>alpha$ implies $g(z)$ is bounded, say $|g(z)|>alpha$.
Then,
$g(z)=1+sum_{n=1}^{infty}a_nz^n \ Rightarrow |g(z)| leq 1+ sum |a_nz^n| $
any hints for showing this.
complex-analysis
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
$endgroup$
Suppose that $g(z)=1+sum_{n=1}^{infty}a_nz^n$ is analytic for $|z|<1$, with $Re(g(z))> alpha$. Show that $ |a_n| leq 2(1-alpha)$ for every $n$.
Answer:
Since $Re(g(z))>alpha$ implies $g(z)$ is bounded, say $|g(z)|>alpha$.
Then,
$g(z)=1+sum_{n=1}^{infty}a_nz^n \ Rightarrow |g(z)| leq 1+ sum |a_nz^n| $
any hints for showing this.
complex-analysis
complex-analysis
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
asked Dec 17 '18 at 4:30
M. A. SARKARM. A. SARKAR
2,1811619
2,1811619
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One method is using Carathéodory lemma, since for
$$dfrac{g(z)-alpha}{1-alpha}=1+sum_{ngeqslant1}dfrac{a_n}{1-alpha}z^n$$
we have ${bf Re}dfrac{g(z)-alpha}{1-alpha}>0$ where $alpha<1$, then $dfrac{g(z)-alpha}{1-alpha}$ is subordinate to $dfrac{1+z}{1-z}$, and Carathéodory lemma says for such function $left|dfrac{a_n}{1-alpha}right|leqslant2$.
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
One method is using Carathéodory lemma, since for
$$dfrac{g(z)-alpha}{1-alpha}=1+sum_{ngeqslant1}dfrac{a_n}{1-alpha}z^n$$
we have ${bf Re}dfrac{g(z)-alpha}{1-alpha}>0$ where $alpha<1$, then $dfrac{g(z)-alpha}{1-alpha}$ is subordinate to $dfrac{1+z}{1-z}$, and Carathéodory lemma says for such function $left|dfrac{a_n}{1-alpha}right|leqslant2$.
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
One method is using Carathéodory lemma, since for
$$dfrac{g(z)-alpha}{1-alpha}=1+sum_{ngeqslant1}dfrac{a_n}{1-alpha}z^n$$
we have ${bf Re}dfrac{g(z)-alpha}{1-alpha}>0$ where $alpha<1$, then $dfrac{g(z)-alpha}{1-alpha}$ is subordinate to $dfrac{1+z}{1-z}$, and Carathéodory lemma says for such function $left|dfrac{a_n}{1-alpha}right|leqslant2$.
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
One method is using Carathéodory lemma, since for
$$dfrac{g(z)-alpha}{1-alpha}=1+sum_{ngeqslant1}dfrac{a_n}{1-alpha}z^n$$
we have ${bf Re}dfrac{g(z)-alpha}{1-alpha}>0$ where $alpha<1$, then $dfrac{g(z)-alpha}{1-alpha}$ is subordinate to $dfrac{1+z}{1-z}$, and Carathéodory lemma says for such function $left|dfrac{a_n}{1-alpha}right|leqslant2$.
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
$endgroup$
One method is using Carathéodory lemma, since for
$$dfrac{g(z)-alpha}{1-alpha}=1+sum_{ngeqslant1}dfrac{a_n}{1-alpha}z^n$$
we have ${bf Re}dfrac{g(z)-alpha}{1-alpha}>0$ where $alpha<1$, then $dfrac{g(z)-alpha}{1-alpha}$ is subordinate to $dfrac{1+z}{1-z}$, and Carathéodory lemma says for such function $left|dfrac{a_n}{1-alpha}right|leqslant2$.
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
answered Dec 17 '18 at 4:50
NosratiNosrati
26.5k62354
26.5k62354
add a comment |
add a comment |
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