If $f(z)=z+a_2z^2+…+a_nz^n$ is injective on the unit disk, then $|a_2|≤frac12(n-1)$












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Suppose $f(z)=z+a_2z^2+...+a_nz^n$ is injective in $D:={z:|z|<1}$. Show that $|a_2|≤(n-1)/2$ and $|a_n|≤1/n$.




Partial Solution



Assume without loss of generality that $a_nne0$. If $f$ is injective, then $f'(z)≠0$ in $D$. Now, $$f'(z)=1+2a_2z+...+na_nz^{n-1}$$ By the fundamental theorem of algebra, $$f'(z)=na_n(z-r_1)...(z-r_{n-1})$$ where $r_1,...,r_{n-1}$ are the roots of $f'(z)$, and $$f'(0)=1=na_n(-1)^nr_1r_2...r_{n-1}$$ Note that $|r_i|geq1$ for each $i=1,...,n-1$, hence $$|r_1r_2...r_{n-1}|geq1$$ and, finally, $$|a_n|=frac1{n,|r_1r_2...r_{n-1}|}leqfrac{1}{n}$$



What I cannot figure out, though, is how to prove that $|a_2|≤(n-1)/2$. I have tried similar arguments, but cannot seem to get it. Any hints would be greatly appreciated.






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    7












    $begingroup$



    Suppose $f(z)=z+a_2z^2+...+a_nz^n$ is injective in $D:={z:|z|<1}$. Show that $|a_2|≤(n-1)/2$ and $|a_n|≤1/n$.




    Partial Solution



    Assume without loss of generality that $a_nne0$. If $f$ is injective, then $f'(z)≠0$ in $D$. Now, $$f'(z)=1+2a_2z+...+na_nz^{n-1}$$ By the fundamental theorem of algebra, $$f'(z)=na_n(z-r_1)...(z-r_{n-1})$$ where $r_1,...,r_{n-1}$ are the roots of $f'(z)$, and $$f'(0)=1=na_n(-1)^nr_1r_2...r_{n-1}$$ Note that $|r_i|geq1$ for each $i=1,...,n-1$, hence $$|r_1r_2...r_{n-1}|geq1$$ and, finally, $$|a_n|=frac1{n,|r_1r_2...r_{n-1}|}leqfrac{1}{n}$$



    What I cannot figure out, though, is how to prove that $|a_2|≤(n-1)/2$. I have tried similar arguments, but cannot seem to get it. Any hints would be greatly appreciated.






    complex-analysis polynomials







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      7


      1



      $begingroup$



      Suppose $f(z)=z+a_2z^2+...+a_nz^n$ is injective in $D:={z:|z|<1}$. Show that $|a_2|≤(n-1)/2$ and $|a_n|≤1/n$.




      Partial Solution



      Assume without loss of generality that $a_nne0$. If $f$ is injective, then $f'(z)≠0$ in $D$. Now, $$f'(z)=1+2a_2z+...+na_nz^{n-1}$$ By the fundamental theorem of algebra, $$f'(z)=na_n(z-r_1)...(z-r_{n-1})$$ where $r_1,...,r_{n-1}$ are the roots of $f'(z)$, and $$f'(0)=1=na_n(-1)^nr_1r_2...r_{n-1}$$ Note that $|r_i|geq1$ for each $i=1,...,n-1$, hence $$|r_1r_2...r_{n-1}|geq1$$ and, finally, $$|a_n|=frac1{n,|r_1r_2...r_{n-1}|}leqfrac{1}{n}$$



      What I cannot figure out, though, is how to prove that $|a_2|≤(n-1)/2$. I have tried similar arguments, but cannot seem to get it. Any hints would be greatly appreciated.






      complex-analysis polynomials







      share|cite|improve this question











      $endgroup$





      Suppose $f(z)=z+a_2z^2+...+a_nz^n$ is injective in $D:={z:|z|<1}$. Show that $|a_2|≤(n-1)/2$ and $|a_n|≤1/n$.




      Partial Solution



      Assume without loss of generality that $a_nne0$. If $f$ is injective, then $f'(z)≠0$ in $D$. Now, $$f'(z)=1+2a_2z+...+na_nz^{n-1}$$ By the fundamental theorem of algebra, $$f'(z)=na_n(z-r_1)...(z-r_{n-1})$$ where $r_1,...,r_{n-1}$ are the roots of $f'(z)$, and $$f'(0)=1=na_n(-1)^nr_1r_2...r_{n-1}$$ Note that $|r_i|geq1$ for each $i=1,...,n-1$, hence $$|r_1r_2...r_{n-1}|geq1$$ and, finally, $$|a_n|=frac1{n,|r_1r_2...r_{n-1}|}leqfrac{1}{n}$$



      What I cannot figure out, though, is how to prove that $|a_2|≤(n-1)/2$. I have tried similar arguments, but cannot seem to get it. Any hints would be greatly appreciated.






      complex-analysis polynomials




      complex-analysis polynomials






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      edited Dec 7 '18 at 19:04









      Did

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      247k23222457










      asked Dec 6 '18 at 23:54









      greeliousgreelious

      19410




      19410






















          1 Answer
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          $begingroup$

          Figured it out. Just to answer my own question: $$f'(z)=1+2a_2z+ ... +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
          $$implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2})$$
          $$begin{align}
          implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2}|\
          & leq nbigl(frac{1}{n}bigr)bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+ ... +|r_1r_2...r_{n-2}(-1)^{n-2}|bigr)\
          end{align}$$
          $$implies |2a_2|leq1+1+ ... +1=n-1$$
          $$implies |a_2|leqfrac{n-1}{2}$$






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          • $begingroup$
            More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
            $endgroup$
            – Did
            Dec 7 '18 at 14:40











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          1 Answer
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          active

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          5












          $begingroup$

          Figured it out. Just to answer my own question: $$f'(z)=1+2a_2z+ ... +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
          $$implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2})$$
          $$begin{align}
          implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2}|\
          & leq nbigl(frac{1}{n}bigr)bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+ ... +|r_1r_2...r_{n-2}(-1)^{n-2}|bigr)\
          end{align}$$
          $$implies |2a_2|leq1+1+ ... +1=n-1$$
          $$implies |a_2|leqfrac{n-1}{2}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
            $endgroup$
            – Did
            Dec 7 '18 at 14:40
















          5












          $begingroup$

          Figured it out. Just to answer my own question: $$f'(z)=1+2a_2z+ ... +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
          $$implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2})$$
          $$begin{align}
          implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2}|\
          & leq nbigl(frac{1}{n}bigr)bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+ ... +|r_1r_2...r_{n-2}(-1)^{n-2}|bigr)\
          end{align}$$
          $$implies |2a_2|leq1+1+ ... +1=n-1$$
          $$implies |a_2|leqfrac{n-1}{2}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
            $endgroup$
            – Did
            Dec 7 '18 at 14:40














          5












          5








          5





          $begingroup$

          Figured it out. Just to answer my own question: $$f'(z)=1+2a_2z+ ... +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
          $$implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2})$$
          $$begin{align}
          implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2}|\
          & leq nbigl(frac{1}{n}bigr)bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+ ... +|r_1r_2...r_{n-2}(-1)^{n-2}|bigr)\
          end{align}$$
          $$implies |2a_2|leq1+1+ ... +1=n-1$$
          $$implies |a_2|leqfrac{n-1}{2}$$






          share|cite|improve this answer











          $endgroup$



          Figured it out. Just to answer my own question: $$f'(z)=1+2a_2z+ ... +na_nz^{n-1}=na_n(z-r_1)...(z-r_{n-1})$$
          $$implies2a_2=na_n(r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2})$$
          $$begin{align}
          implies|2a_2| & = n|a_n||r_2r_3...r_{n-1}(-1)^{n-2}+r_1r_3...r_{n-1}(-1)^{n-2}+ ... +r_1r_2...r_{n-2}(-1)^{n-2}|\
          & leq nbigl(frac{1}{n}bigr)bigl(|r_2r_3...r_{n-1}(-1)^{n-2}|+|r_1r_3...r_{n-1}(-1)^{n-2}|+ ... +|r_1r_2...r_{n-2}(-1)^{n-2}|bigr)\
          end{align}$$
          $$implies |2a_2|leq1+1+ ... +1=n-1$$
          $$implies |a_2|leqfrac{n-1}{2}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 14:07

























          answered Dec 7 '18 at 0:14









          greeliousgreelious

          19410




          19410












          • $begingroup$
            More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
            $endgroup$
            – Did
            Dec 7 '18 at 14:40


















          • $begingroup$
            More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
            $endgroup$
            – Did
            Dec 7 '18 at 14:40
















          $begingroup$
          More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
          $endgroup$
          – Did
          Dec 7 '18 at 14:40




          $begingroup$
          More simply, $$2a_2=-sum_{k=1}^{n-1}frac1{r_k}$$
          $endgroup$
          – Did
          Dec 7 '18 at 14:40


















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