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nth roots and logarithm

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3

$begingroup$

Suppose $Omega$ is a region,$fin H(Omega)$, and $fnotequiv 0$.If for every positive integer n,there is $f_n in H(Omega)$,$(f_n)^n=f$,prove that there is $gin H(Omega),f=e^g$.

My conjecture :

$$g=lim (nf_n - n)$$

But I don't know how to deal with it.

I will appreciate your help.

complex-analysis

share|cite|improve this question

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can prove convergence, then that should be ok. But here it is sufficient to show that $g=ln(f)$ is well-defined, i.e., that $f$ does not take values on the negative half axis.
  $endgroup$
  – LutzL
  Mar 20 '14 at 14:53
  

  
  
  
  

add a comment | 

3

$begingroup$

Suppose $Omega$ is a region,$fin H(Omega)$, and $fnotequiv 0$.If for every positive integer n,there is $f_n in H(Omega)$,$(f_n)^n=f$,prove that there is $gin H(Omega),f=e^g$.

My conjecture :

$$g=lim (nf_n - n)$$

But I don't know how to deal with it.

I will appreciate your help.

complex-analysis

share|cite|improve this question

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can prove convergence, then that should be ok. But here it is sufficient to show that $g=ln(f)$ is well-defined, i.e., that $f$ does not take values on the negative half axis.
  $endgroup$
  – LutzL
  Mar 20 '14 at 14:53
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $Omega$ is a region,$fin H(Omega)$, and $fnotequiv 0$.If for every positive integer n,there is $f_n in H(Omega)$,$(f_n)^n=f$,prove that there is $gin H(Omega),f=e^g$.

My conjecture :

$$g=lim (nf_n - n)$$

But I don't know how to deal with it.

I will appreciate your help.

complex-analysis

share|cite|improve this question

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

$endgroup$

share|cite|improve this question

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

share|cite|improve this question

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

asked Mar 20 '14 at 3:05

gilliattgilliatt

1478

1 Answer
1

active

oldest

votes

5

$begingroup$

If $f$ has a holomorphic logarithm $g$, then $f$ has no zeros, and $f' = g'cdot e^g$, or

$$g' = fracf'f,$$

so the function $f'/f$ has a primitive on $Omega$. Conversely, if $f$ has no zeros, and $f'/f$ has a primitive $h$ on $Omega$, then $fcdot e^-h$ is constant, and by adding a suitable constant to $h$, we obtain a $g$ with $f = e^g$.

So first, we deduce that $f$ has no zeros in $Omega$: If $f$ had a zero in $a in Omega$, say of order $k$, then the $n$-th root $f_n$ would also have have a zero in $a$. Let its order be $k_n$, so $f_n(z) = (z-a)^k_ncdot h_n(z)$ with $h_n in H(Omega)$ such that $h_n(a) neq 0$. But then $f(z) = (z-a)^ncdot k_nh_n(z)^n$, so $k = ncdot k_n$, and $n mid k$. But a positive integer has only finitely many divisors, so $f$ can't have a zero.

Thus $q = f'/f in H(Omega)$, and $q$ has a primitive if and only if

$$I(gamma) := int_gamma fracf'(z)f(z),dz = 0$$

for all closed paths $gamma$ in $Omega$. But $n(gamma) = frac12pi iI(gamma)$ is the winding number of the closed path $fcirc gamma$ around $0$, hence an integer, and

$$beginalign
n(gamma) &= frac12pi iint_gamma fracf'(z)f(z),dz\
&= frac12pi i int_gamma frac(f_n(z)^n)'f_n(z)^n,dz\
&= frac12pi iint_gamma fracn f_n(z)^n-1f_n'(z)f_n(z)^n,dz\
&= fracn2pi iint_gamma fracf_n'(z)f_n(z),dz
endalign$$

is $n$ times the winding number of the closed path $f_ncircgamma$ around $0$, hence a multiple of $n$. The only integer that is a multiply of all $n in mathbbZ^+$ is $0$, hence $n(gamma) = 0$, and indeed $f'/f$ has a primitive on $Omega$.

share|cite|improve this answer

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

$endgroup$

add a comment | 

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5

$begingroup$

If $f$ has a holomorphic logarithm $g$, then $f$ has no zeros, and $f' = g'cdot e^g$, or

$$g' = fracf'f,$$

so the function $f'/f$ has a primitive on $Omega$. Conversely, if $f$ has no zeros, and $f'/f$ has a primitive $h$ on $Omega$, then $fcdot e^-h$ is constant, and by adding a suitable constant to $h$, we obtain a $g$ with $f = e^g$.

So first, we deduce that $f$ has no zeros in $Omega$: If $f$ had a zero in $a in Omega$, say of order $k$, then the $n$-th root $f_n$ would also have have a zero in $a$. Let its order be $k_n$, so $f_n(z) = (z-a)^k_ncdot h_n(z)$ with $h_n in H(Omega)$ such that $h_n(a) neq 0$. But then $f(z) = (z-a)^ncdot k_nh_n(z)^n$, so $k = ncdot k_n$, and $n mid k$. But a positive integer has only finitely many divisors, so $f$ can't have a zero.

Thus $q = f'/f in H(Omega)$, and $q$ has a primitive if and only if

$$I(gamma) := int_gamma fracf'(z)f(z),dz = 0$$

for all closed paths $gamma$ in $Omega$. But $n(gamma) = frac12pi iI(gamma)$ is the winding number of the closed path $fcirc gamma$ around $0$, hence an integer, and

$$beginalign
n(gamma) &= frac12pi iint_gamma fracf'(z)f(z),dz\
&= frac12pi i int_gamma frac(f_n(z)^n)'f_n(z)^n,dz\
&= frac12pi iint_gamma fracn f_n(z)^n-1f_n'(z)f_n(z)^n,dz\
&= fracn2pi iint_gamma fracf_n'(z)f_n(z),dz
endalign$$

is $n$ times the winding number of the closed path $f_ncircgamma$ around $0$, hence a multiple of $n$. The only integer that is a multiply of all $n in mathbbZ^+$ is $0$, hence $n(gamma) = 0$, and indeed $f'/f$ has a primitive on $Omega$.

share|cite|improve this answer

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

$endgroup$

add a comment | 

5

$begingroup$

If $f$ has a holomorphic logarithm $g$, then $f$ has no zeros, and $f' = g'cdot e^g$, or

$$g' = fracf'f,$$

so the function $f'/f$ has a primitive on $Omega$. Conversely, if $f$ has no zeros, and $f'/f$ has a primitive $h$ on $Omega$, then $fcdot e^-h$ is constant, and by adding a suitable constant to $h$, we obtain a $g$ with $f = e^g$.

So first, we deduce that $f$ has no zeros in $Omega$: If $f$ had a zero in $a in Omega$, say of order $k$, then the $n$-th root $f_n$ would also have have a zero in $a$. Let its order be $k_n$, so $f_n(z) = (z-a)^k_ncdot h_n(z)$ with $h_n in H(Omega)$ such that $h_n(a) neq 0$. But then $f(z) = (z-a)^ncdot k_nh_n(z)^n$, so $k = ncdot k_n$, and $n mid k$. But a positive integer has only finitely many divisors, so $f$ can't have a zero.

Thus $q = f'/f in H(Omega)$, and $q$ has a primitive if and only if

$$I(gamma) := int_gamma fracf'(z)f(z),dz = 0$$

for all closed paths $gamma$ in $Omega$. But $n(gamma) = frac12pi iI(gamma)$ is the winding number of the closed path $fcirc gamma$ around $0$, hence an integer, and

$$beginalign
n(gamma) &= frac12pi iint_gamma fracf'(z)f(z),dz\
&= frac12pi i int_gamma frac(f_n(z)^n)'f_n(z)^n,dz\
&= frac12pi iint_gamma fracn f_n(z)^n-1f_n'(z)f_n(z)^n,dz\
&= fracn2pi iint_gamma fracf_n'(z)f_n(z),dz
endalign$$

is $n$ times the winding number of the closed path $f_ncircgamma$ around $0$, hence a multiple of $n$. The only integer that is a multiply of all $n in mathbbZ^+$ is $0$, hence $n(gamma) = 0$, and indeed $f'/f$ has a primitive on $Omega$.

share|cite|improve this answer

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

$endgroup$

add a comment | 

$begingroup$

If $f$ has a holomorphic logarithm $g$, then $f$ has no zeros, and $f' = g'cdot e^g$, or

$$g' = fracf'f,$$

so the function $f'/f$ has a primitive on $Omega$. Conversely, if $f$ has no zeros, and $f'/f$ has a primitive $h$ on $Omega$, then $fcdot e^-h$ is constant, and by adding a suitable constant to $h$, we obtain a $g$ with $f = e^g$.

So first, we deduce that $f$ has no zeros in $Omega$: If $f$ had a zero in $a in Omega$, say of order $k$, then the $n$-th root $f_n$ would also have have a zero in $a$. Let its order be $k_n$, so $f_n(z) = (z-a)^k_ncdot h_n(z)$ with $h_n in H(Omega)$ such that $h_n(a) neq 0$. But then $f(z) = (z-a)^ncdot k_nh_n(z)^n$, so $k = ncdot k_n$, and $n mid k$. But a positive integer has only finitely many divisors, so $f$ can't have a zero.

Thus $q = f'/f in H(Omega)$, and $q$ has a primitive if and only if

$$I(gamma) := int_gamma fracf'(z)f(z),dz = 0$$

for all closed paths $gamma$ in $Omega$. But $n(gamma) = frac12pi iI(gamma)$ is the winding number of the closed path $fcirc gamma$ around $0$, hence an integer, and

$$beginalign
n(gamma) &= frac12pi iint_gamma fracf'(z)f(z),dz\
&= frac12pi i int_gamma frac(f_n(z)^n)'f_n(z)^n,dz\
&= frac12pi iint_gamma fracn f_n(z)^n-1f_n'(z)f_n(z)^n,dz\
&= fracn2pi iint_gamma fracf_n'(z)f_n(z),dz
endalign$$

is $n$ times the winding number of the closed path $f_ncircgamma$ around $0$, hence a multiple of $n$. The only integer that is a multiply of all $n in mathbbZ^+$ is $0$, hence $n(gamma) = 0$, and indeed $f'/f$ has a primitive on $Omega$.

share|cite|improve this answer

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

$endgroup$

share|cite|improve this answer

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289

answered Mar 21 '14 at 13:43

Daniel FischerDaniel Fischer

174k17171289