Complex Polynomial Inequality Proof











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I'm trying to solve this question but I am having trouble connecting the dots. The question reads:




Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.




So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.



Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?










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  • 1




    Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
    – Gerry Myerson
    Nov 12 at 4:55















up vote
2
down vote

favorite












I'm trying to solve this question but I am having trouble connecting the dots. The question reads:




Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.




So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.



Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?










share|cite|improve this question




















  • 1




    Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
    – Gerry Myerson
    Nov 12 at 4:55













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I'm trying to solve this question but I am having trouble connecting the dots. The question reads:




Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.




So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.



Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?










share|cite|improve this question















I'm trying to solve this question but I am having trouble connecting the dots. The question reads:




Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.




So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.



Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?







sequences-and-series complex-analysis






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edited Nov 12 at 5:10









Wesley Strik

1,378322




1,378322










asked Nov 12 at 4:25









Felicio Grande

499619




499619








  • 1




    Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
    – Gerry Myerson
    Nov 12 at 4:55














  • 1




    Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
    – Gerry Myerson
    Nov 12 at 4:55








1




1




Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55




Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55










2 Answers
2






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4
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This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.






share|cite|improve this answer





















  • Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
    – Gerry Myerson
    Nov 13 at 10:06


















up vote
2
down vote













One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$

Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$

which gives $|a_n|leq 1$.






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    2 Answers
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    active

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    2 Answers
    2






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    up vote
    4
    down vote













    This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.






    share|cite|improve this answer





















    • Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
      – Gerry Myerson
      Nov 13 at 10:06















    up vote
    4
    down vote













    This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.






    share|cite|improve this answer





















    • Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
      – Gerry Myerson
      Nov 13 at 10:06













    up vote
    4
    down vote










    up vote
    4
    down vote









    This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.






    share|cite|improve this answer












    This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 12 at 5:31









    Kavi Rama Murthy

    46.6k31854




    46.6k31854












    • Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
      – Gerry Myerson
      Nov 13 at 10:06


















    • Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
      – Gerry Myerson
      Nov 13 at 10:06
















    Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
    – Gerry Myerson
    Nov 13 at 10:06




    Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
    – Gerry Myerson
    Nov 13 at 10:06










    up vote
    2
    down vote













    One way to show this is via Cauchy estimates.
    From Cauchy integral formula, for the $n$-th derivative of $P$ we have
    $$
    P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
    $$

    Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
    $$
    n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
    frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
    $$

    which gives $|a_n|leq 1$.






    share|cite|improve this answer

























      up vote
      2
      down vote













      One way to show this is via Cauchy estimates.
      From Cauchy integral formula, for the $n$-th derivative of $P$ we have
      $$
      P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
      $$

      Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
      $$
      n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
      frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
      $$

      which gives $|a_n|leq 1$.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        One way to show this is via Cauchy estimates.
        From Cauchy integral formula, for the $n$-th derivative of $P$ we have
        $$
        P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
        $$

        Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
        $$
        n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
        frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
        $$

        which gives $|a_n|leq 1$.






        share|cite|improve this answer












        One way to show this is via Cauchy estimates.
        From Cauchy integral formula, for the $n$-th derivative of $P$ we have
        $$
        P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
        $$

        Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
        $$
        n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
        frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
        $$

        which gives $|a_n|leq 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 12 at 5:04









        Hayk

        2,022212




        2,022212






























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