6 true or false questions in complex analysis












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$begingroup$


Let $f:mathbb Ctomathbb C$ be an entire function. Which of the following statements are true and which are false?




  1. If $f(z)inmathbb R$ for all $zinmathbb C$, then $f$ is constant.

  2. If $f(frac{1}{n})=frac{i}{n}$ for all $ninmathbb N$, then $f(z)=iz$ for all $zinmathbb C$.

  3. If $f$ is a non-constant polynomial, then there is a continuously differentiable path $gamma: [0,1]tomathbb C$ such that $int_gamma f(z)dz=2pi i$.

  4. The function $1/f$ has no pole in $0$.

  5. The function $rmapstoint_{|z|=r}f(z)dz$ is constant on $(0,infty)$.

  6. The series $sum_{n=0}^inftyfrac{1}{n!}f^{(n)}(1)(z-1)^n$ converges for all $zinmathbb C$



Here is what I got:




  1. True. If $f$ is not constant, then $f(mathbb C)$ must be a domain which conflicts with $f(mathbb C)subseteqmathbb R$.

  2. True. On the set ${1/n|ninmathbb N$} the function $f$ coincides with $g(z)=iz$. Since the set does have an accumulation point in the domain of $f$, the identity theorem yields $f=g$.

  3. I have no idea. How do I start here?

  4. False. $f(z)=z$.

  5. True. Since $f$ is holomorphic, the integral is always $0$.

  6. True. This is just the taylor series of $f$ in $z_0=1$.






complex-analysis proof-verification entire-functions







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$endgroup$

















    0












    $begingroup$


    Let $f:mathbb Ctomathbb C$ be an entire function. Which of the following statements are true and which are false?




    1. If $f(z)inmathbb R$ for all $zinmathbb C$, then $f$ is constant.

    2. If $f(frac{1}{n})=frac{i}{n}$ for all $ninmathbb N$, then $f(z)=iz$ for all $zinmathbb C$.

    3. If $f$ is a non-constant polynomial, then there is a continuously differentiable path $gamma: [0,1]tomathbb C$ such that $int_gamma f(z)dz=2pi i$.

    4. The function $1/f$ has no pole in $0$.

    5. The function $rmapstoint_{|z|=r}f(z)dz$ is constant on $(0,infty)$.

    6. The series $sum_{n=0}^inftyfrac{1}{n!}f^{(n)}(1)(z-1)^n$ converges for all $zinmathbb C$



    Here is what I got:




    1. True. If $f$ is not constant, then $f(mathbb C)$ must be a domain which conflicts with $f(mathbb C)subseteqmathbb R$.

    2. True. On the set ${1/n|ninmathbb N$} the function $f$ coincides with $g(z)=iz$. Since the set does have an accumulation point in the domain of $f$, the identity theorem yields $f=g$.

    3. I have no idea. How do I start here?

    4. False. $f(z)=z$.

    5. True. Since $f$ is holomorphic, the integral is always $0$.

    6. True. This is just the taylor series of $f$ in $z_0=1$.






    complex-analysis proof-verification entire-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f:mathbb Ctomathbb C$ be an entire function. Which of the following statements are true and which are false?




      1. If $f(z)inmathbb R$ for all $zinmathbb C$, then $f$ is constant.

      2. If $f(frac{1}{n})=frac{i}{n}$ for all $ninmathbb N$, then $f(z)=iz$ for all $zinmathbb C$.

      3. If $f$ is a non-constant polynomial, then there is a continuously differentiable path $gamma: [0,1]tomathbb C$ such that $int_gamma f(z)dz=2pi i$.

      4. The function $1/f$ has no pole in $0$.

      5. The function $rmapstoint_{|z|=r}f(z)dz$ is constant on $(0,infty)$.

      6. The series $sum_{n=0}^inftyfrac{1}{n!}f^{(n)}(1)(z-1)^n$ converges for all $zinmathbb C$



      Here is what I got:




      1. True. If $f$ is not constant, then $f(mathbb C)$ must be a domain which conflicts with $f(mathbb C)subseteqmathbb R$.

      2. True. On the set ${1/n|ninmathbb N$} the function $f$ coincides with $g(z)=iz$. Since the set does have an accumulation point in the domain of $f$, the identity theorem yields $f=g$.

      3. I have no idea. How do I start here?

      4. False. $f(z)=z$.

      5. True. Since $f$ is holomorphic, the integral is always $0$.

      6. True. This is just the taylor series of $f$ in $z_0=1$.






      complex-analysis proof-verification entire-functions







      share|cite|improve this question











      $endgroup$




      Let $f:mathbb Ctomathbb C$ be an entire function. Which of the following statements are true and which are false?




      1. If $f(z)inmathbb R$ for all $zinmathbb C$, then $f$ is constant.

      2. If $f(frac{1}{n})=frac{i}{n}$ for all $ninmathbb N$, then $f(z)=iz$ for all $zinmathbb C$.

      3. If $f$ is a non-constant polynomial, then there is a continuously differentiable path $gamma: [0,1]tomathbb C$ such that $int_gamma f(z)dz=2pi i$.

      4. The function $1/f$ has no pole in $0$.

      5. The function $rmapstoint_{|z|=r}f(z)dz$ is constant on $(0,infty)$.

      6. The series $sum_{n=0}^inftyfrac{1}{n!}f^{(n)}(1)(z-1)^n$ converges for all $zinmathbb C$



      Here is what I got:




      1. True. If $f$ is not constant, then $f(mathbb C)$ must be a domain which conflicts with $f(mathbb C)subseteqmathbb R$.

      2. True. On the set ${1/n|ninmathbb N$} the function $f$ coincides with $g(z)=iz$. Since the set does have an accumulation point in the domain of $f$, the identity theorem yields $f=g$.

      3. I have no idea. How do I start here?

      4. False. $f(z)=z$.

      5. True. Since $f$ is holomorphic, the integral is always $0$.

      6. True. This is just the taylor series of $f$ in $z_0=1$.






      complex-analysis proof-verification entire-functions




      complex-analysis proof-verification entire-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 12:20









      José Carlos Santos

      172k23132240




      172k23132240










      asked Dec 19 '18 at 12:14









      RedLanternRedLantern

      517




      517






















          1 Answer
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          1












          $begingroup$

          What you did is correct.



          The assertion 3. is true. For each $winmathbb C$, let $gamma_wcolon[0,1]longrightarrowmathbb C$ be the path $gamma_w(t)=tw$. Then the map $wmapstoint_{gamma_w}f(z),mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2pi i$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
            $endgroup$
            – RedLantern
            Dec 19 '18 at 12:35






          • 1




            $begingroup$
            By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
            $endgroup$
            – Fred
            Dec 19 '18 at 12:54












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          active

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          1












          $begingroup$

          What you did is correct.



          The assertion 3. is true. For each $winmathbb C$, let $gamma_wcolon[0,1]longrightarrowmathbb C$ be the path $gamma_w(t)=tw$. Then the map $wmapstoint_{gamma_w}f(z),mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2pi i$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
            $endgroup$
            – RedLantern
            Dec 19 '18 at 12:35






          • 1




            $begingroup$
            By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
            $endgroup$
            – Fred
            Dec 19 '18 at 12:54
















          1












          $begingroup$

          What you did is correct.



          The assertion 3. is true. For each $winmathbb C$, let $gamma_wcolon[0,1]longrightarrowmathbb C$ be the path $gamma_w(t)=tw$. Then the map $wmapstoint_{gamma_w}f(z),mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2pi i$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
            $endgroup$
            – RedLantern
            Dec 19 '18 at 12:35






          • 1




            $begingroup$
            By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
            $endgroup$
            – Fred
            Dec 19 '18 at 12:54














          1












          1








          1





          $begingroup$

          What you did is correct.



          The assertion 3. is true. For each $winmathbb C$, let $gamma_wcolon[0,1]longrightarrowmathbb C$ be the path $gamma_w(t)=tw$. Then the map $wmapstoint_{gamma_w}f(z),mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2pi i$.






          share|cite|improve this answer









          $endgroup$



          What you did is correct.



          The assertion 3. is true. For each $winmathbb C$, let $gamma_wcolon[0,1]longrightarrowmathbb C$ be the path $gamma_w(t)=tw$. Then the map $wmapstoint_{gamma_w}f(z),mathrm dz$ is just the antiderivative $F$ of $f$ such that $F(0)=0$. But then $F$ is a non constant polynomial function and therefore it is surjective. So, there is so $w$ such that $F(w)=2pi i$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 12:19









          José Carlos SantosJosé Carlos Santos

          172k23132240




          172k23132240












          • $begingroup$
            So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
            $endgroup$
            – RedLantern
            Dec 19 '18 at 12:35






          • 1




            $begingroup$
            By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
            $endgroup$
            – Fred
            Dec 19 '18 at 12:54


















          • $begingroup$
            So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
            $endgroup$
            – RedLantern
            Dec 19 '18 at 12:35






          • 1




            $begingroup$
            By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
            $endgroup$
            – Fred
            Dec 19 '18 at 12:54
















          $begingroup$
          So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
          $endgroup$
          – RedLantern
          Dec 19 '18 at 12:35




          $begingroup$
          So all non constant polynomials are surjective? I didn't know that. And what about injective? Is the proof rather easy or complicated?
          $endgroup$
          – RedLantern
          Dec 19 '18 at 12:35




          1




          1




          $begingroup$
          By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
          $endgroup$
          – Fred
          Dec 19 '18 at 12:54




          $begingroup$
          By the fundamental theorem of algebra, a non constant polynomial is surjective. In general, a polynomial is not injective. Example: $p(z)=z^2$.
          $endgroup$
          – Fred
          Dec 19 '18 at 12:54


















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