Construct an analytic function with zeros at intersection of zeros of other analytic functions












1












$begingroup$


Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$



In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!






complex-analysis roots







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If you are asking for a simple formula for $h$ I doubt if such a thing exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 6:07












  • $begingroup$
    For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
    $endgroup$
    – Arthur
    Jan 7 at 6:41










  • $begingroup$
    @KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 7:29












  • $begingroup$
    @JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 7:37










  • $begingroup$
    If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
    $endgroup$
    – reuns
    Jan 7 at 8:08
















1












$begingroup$


Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$



In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!






complex-analysis roots







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If you are asking for a simple formula for $h$ I doubt if such a thing exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 6:07












  • $begingroup$
    For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
    $endgroup$
    – Arthur
    Jan 7 at 6:41










  • $begingroup$
    @KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 7:29












  • $begingroup$
    @JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 7:37










  • $begingroup$
    If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
    $endgroup$
    – reuns
    Jan 7 at 8:08














1












1








1


0



$begingroup$


Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$



In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!






complex-analysis roots







share|cite|improve this question











$endgroup$




Given two complex analytic functions f and g on some compact domain $D$, I want to construct a new analytic function h on $D$ such that
$$h(z) = 0 iff f(z) = 0 text{ and } g(z) = 0$$
I want to do this without finding the zeros of $f$ and $g$.
If I wanted to find $h$ such that
$$
h(z) = 0 iff f(z) = 0 text{ or } g(z) = 0,
$$
this can be easily done by letting
$$
h(z) = f(z)g(z).
$$



In a more specific version, we can assume that zeros of $f$ and $g$ lie on the real line.
Thanks!






complex-analysis roots




complex-analysis roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 7:53









Daniele Tampieri

1,9841619




1,9841619










asked Jan 7 at 5:43









Jaeyoon KimJaeyoon Kim

84




84








  • 2




    $begingroup$
    If you are asking for a simple formula for $h$ I doubt if such a thing exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 6:07












  • $begingroup$
    For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
    $endgroup$
    – Arthur
    Jan 7 at 6:41










  • $begingroup$
    @KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 7:29












  • $begingroup$
    @JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 7:37










  • $begingroup$
    If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
    $endgroup$
    – reuns
    Jan 7 at 8:08














  • 2




    $begingroup$
    If you are asking for a simple formula for $h$ I doubt if such a thing exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 6:07












  • $begingroup$
    For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
    $endgroup$
    – Arthur
    Jan 7 at 6:41










  • $begingroup$
    @KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 7:29












  • $begingroup$
    @JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 7:37










  • $begingroup$
    If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
    $endgroup$
    – reuns
    Jan 7 at 8:08








2




2




$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07






$begingroup$
If you are asking for a simple formula for $h$ I doubt if such a thing exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 6:07














$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41




$begingroup$
For polynomials it's just the greatest common denominator. For general analytic functions I have no idea.
$endgroup$
– Arthur
Jan 7 at 6:41












$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29






$begingroup$
@KaviRamaMurthy such $h$ does exists. If we assume that $f$ and $g$ are not identically zero, then by isolated zeros, we know that there are finitely many zeros. Let $z_1,ldots,z_n$ be all the values such that $f(z_i)$ and $g(z_i)$ are zero. Let $h(z) = prod_{i}(z-z_i)$. But yes, I don't see how to easily compute for $h$.
$endgroup$
– Jaeyoon Kim
Jan 7 at 7:29














$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37




$begingroup$
@JaeyoonKim Surely, $h$ exists with the desired property. But you appear to seek a simple formula for $h$ and I suspect that no such formula exists.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 7:37












$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08




$begingroup$
If the zeros are simple you may try to find $u,v$ analytic such that $w(z)=frac{f'(z)}{f(z)}+e^{u(z)} frac{g'(z)}{g(z)}+v(z)$ has no zeros, so $f(z)g(z)w(z)$ is what you want. Asking if $u,v$ always exist is an interesting problem.
$endgroup$
– reuns
Jan 7 at 8:08










1 Answer
1






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$begingroup$

There is no formula of the form $$h(z)=F(f(z),g(z)).$$



Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 18:14










  • $begingroup$
    @JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
    $endgroup$
    – David C. Ullrich
    Jan 7 at 18:22













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$begingroup$

There is no formula of the form $$h(z)=F(f(z),g(z)).$$



Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 18:14










  • $begingroup$
    @JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
    $endgroup$
    – David C. Ullrich
    Jan 7 at 18:22


















0












$begingroup$

There is no formula of the form $$h(z)=F(f(z),g(z)).$$



Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 18:14










  • $begingroup$
    @JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
    $endgroup$
    – David C. Ullrich
    Jan 7 at 18:22
















0












0








0





$begingroup$

There is no formula of the form $$h(z)=F(f(z),g(z)).$$



Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.






share|cite|improve this answer









$endgroup$



There is no formula of the form $$h(z)=F(f(z),g(z)).$$



Proof: Letting $f(z)=z$, $g(z)=c$ shows that $zmapsto F(z,c)$ is holomorphic. Similarly for $F(c,w)$, so a theorem of Hartogs shows that $F$ is a holomorphic function of two variables. Now $F(0,0)=0$, but the zeroes of a holomorphic function in $Bbb C^2$ cannot be isolated, by another theorem of Hartogs. So it cannot happen that $F(f(z),g(z))=0$ if and only if $f(z)=g(z)=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 16:10









David C. UllrichDavid C. Ullrich

59.9k43994




59.9k43994












  • $begingroup$
    Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 18:14










  • $begingroup$
    @JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
    $endgroup$
    – David C. Ullrich
    Jan 7 at 18:22




















  • $begingroup$
    Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
    $endgroup$
    – Jaeyoon Kim
    Jan 7 at 18:14










  • $begingroup$
    @JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
    $endgroup$
    – David C. Ullrich
    Jan 7 at 18:22


















$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14




$begingroup$
Thank you for your help! I am actually asking for a weaker condition. I don't require $F$ to be holomorphic.
$endgroup$
– Jaeyoon Kim
Jan 7 at 18:14












$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22






$begingroup$
@JaeyoonKim ??? I didn't require $F$ to be holomorphic! If $F(f,g)$ is a holomorphic function with zero set the intersection of the two zero sets, for eery $f$ and $g$, it follows that $F$ is holomorphic - I gave a proof of that.
$endgroup$
– David C. Ullrich
Jan 7 at 18:22




















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