Mixing
Square Root of a Holomorphic Function
Clash Royale CLAN TAG#URR8PPP
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Let $U$ be an open connected subset of $mathbbC$ and $nin mathbbN$ be $even$. Let $z_1$, $z_2$,....$z_n$ be $n$ distinct complex numbers lying in the same connected component of $mathbbC-U$. Show that there exists a holomorphic function $f$ on $U$ such that $(f(z))^2$=$(z-z_1)....(z-z_n)$ $forall z in U$.
I have managed to prove this result if $n=2$, but I am not able to genaralize it. Thanks for any help.
complex-analysis
asked Jul 15 at 13:03
Ester
8791925
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show 1 more comment
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0
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favorite
Let $U$ be an open connected subset of $mathbbC$ and $nin mathbbN$ be $even$. Let $z_1$, $z_2$,....$z_n$ be $n$ distinct complex numbers lying in the same connected component of $mathbbC-U$. Show that there exists a holomorphic function $f$ on $U$ such that $(f(z))^2$=$(z-z_1)....(z-z_n)$ $forall z in U$.
I have managed to prove this result if $n=2$, but I am not able to genaralize it. Thanks for any help.
complex-analysis
asked Jul 15 at 13:03
Ester
8791925
2
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $U$ be an open connected subset of $mathbbC$ and $nin mathbbN$ be $even$. Let $z_1$, $z_2$,....$z_n$ be $n$ distinct complex numbers lying in the same connected component of $mathbbC-U$. Show that there exists a holomorphic function $f$ on $U$ such that $(f(z))^2$=$(z-z_1)....(z-z_n)$ $forall z in U$.
I have managed to prove this result if $n=2$, but I am not able to genaralize it. Thanks for any help.
complex-analysis
asked Jul 15 at 13:03
Ester
8791925
Let $U$ be an open connected subset of $mathbbC$ and $nin mathbbN$ be $even$. Let $z_1$, $z_2$,....$z_n$ be $n$ distinct complex numbers lying in the same connected component of $mathbbC-U$. Show that there exists a holomorphic function $f$ on $U$ such that $(f(z))^2$=$(z-z_1)....(z-z_n)$ $forall z in U$.
I have managed to prove this result if $n=2$, but I am not able to genaralize it. Thanks for any help.
complex-analysis
asked Jul 15 at 13:03
Ester
8791925
asked Jul 15 at 13:03
Ester
8791925
asked Jul 15 at 13:03
Ester
8791925
asked Jul 15 at 13:03
Ester
8791925
8791925
2
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22
 |Â
show 1 more comment
2
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22
2
2
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of $(z - z_2k-1)(z - z_2k)$, $k = 1, ldots, frac n2$.
Alternatively, you can show that if $z_1, ldots, z_n$ lie in the same
component of $Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ tag*
g(z)^n = (z-z_1) cdots (z-z_n) , .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^n/2$.
To show the existence of $g$, note that for any closed curve
$gamma$ in $U$ the winding number $N(gamma, z_i)$
is independent of $i$, so that
$$
frac 1n int_gamma left( frac1z-z_1 + ldots + frac1z-z_nright) , dz = N(gamma, z_1)
$$
is a multiple of $2 pi i$. Therefore we can choose $z_0 in U$ and define
$$
g(z) = exp left( frac 1n int_z_0^z left( frac1w-z_1 + ldots + frac1w-z_nright) , dw right) , .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
h(z) = g(z)^-n (z-z_1)cdots(z-z_n)
$$
satisfies
$$
frach'(z)h(z) = -n fracg'(z)g(z) + frac1z-z_1 + ldots + frac1z-z_n = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
answered Jul 15 at 16:18
Martin R
23.9k32743
add a comment |Â
up vote
0
down vote
Assume that $z_2k-1$ lies in the same connected component of $Bbb Csetminus U$ as $z_2k$. Let $w_k(t)$ be a path that connects $w_k(0)=z_2k-1$ with $w_k(1)=z_2k$. Then the homotopy
$$
h(t,z)=prod_k=0^n/2(z-z_2k-1)(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=prod_k=0^n/2(z-z_2k-1)
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
answered Jul 15 at 17:25
LutzL
49.8k31849
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of $(z - z_2k-1)(z - z_2k)$, $k = 1, ldots, frac n2$.
Alternatively, you can show that if $z_1, ldots, z_n$ lie in the same
component of $Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ tag*
g(z)^n = (z-z_1) cdots (z-z_n) , .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^n/2$.
To show the existence of $g$, note that for any closed curve
$gamma$ in $U$ the winding number $N(gamma, z_i)$
is independent of $i$, so that
$$
frac 1n int_gamma left( frac1z-z_1 + ldots + frac1z-z_nright) , dz = N(gamma, z_1)
$$
is a multiple of $2 pi i$. Therefore we can choose $z_0 in U$ and define
$$
g(z) = exp left( frac 1n int_z_0^z left( frac1w-z_1 + ldots + frac1w-z_nright) , dw right) , .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
h(z) = g(z)^-n (z-z_1)cdots(z-z_n)
$$
satisfies
$$
frach'(z)h(z) = -n fracg'(z)g(z) + frac1z-z_1 + ldots + frac1z-z_n = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
answered Jul 15 at 16:18
Martin R
23.9k32743
add a comment |Â
up vote
2
down vote
accepted
As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of $(z - z_2k-1)(z - z_2k)$, $k = 1, ldots, frac n2$.
Alternatively, you can show that if $z_1, ldots, z_n$ lie in the same
component of $Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ tag*
g(z)^n = (z-z_1) cdots (z-z_n) , .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^n/2$.
To show the existence of $g$, note that for any closed curve
$gamma$ in $U$ the winding number $N(gamma, z_i)$
is independent of $i$, so that
$$
frac 1n int_gamma left( frac1z-z_1 + ldots + frac1z-z_nright) , dz = N(gamma, z_1)
$$
is a multiple of $2 pi i$. Therefore we can choose $z_0 in U$ and define
$$
g(z) = exp left( frac 1n int_z_0^z left( frac1w-z_1 + ldots + frac1w-z_nright) , dw right) , .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
h(z) = g(z)^-n (z-z_1)cdots(z-z_n)
$$
satisfies
$$
frach'(z)h(z) = -n fracg'(z)g(z) + frac1z-z_1 + ldots + frac1z-z_n = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
answered Jul 15 at 16:18
Martin R
23.9k32743
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of $(z - z_2k-1)(z - z_2k)$, $k = 1, ldots, frac n2$.
Alternatively, you can show that if $z_1, ldots, z_n$ lie in the same
component of $Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ tag*
g(z)^n = (z-z_1) cdots (z-z_n) , .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^n/2$.
To show the existence of $g$, note that for any closed curve
$gamma$ in $U$ the winding number $N(gamma, z_i)$
is independent of $i$, so that
$$
frac 1n int_gamma left( frac1z-z_1 + ldots + frac1z-z_nright) , dz = N(gamma, z_1)
$$
is a multiple of $2 pi i$. Therefore we can choose $z_0 in U$ and define
$$
g(z) = exp left( frac 1n int_z_0^z left( frac1w-z_1 + ldots + frac1w-z_nright) , dw right) , .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
h(z) = g(z)^-n (z-z_1)cdots(z-z_n)
$$
satisfies
$$
frach'(z)h(z) = -n fracg'(z)g(z) + frac1z-z_1 + ldots + frac1z-z_n = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
answered Jul 15 at 16:18
Martin R
23.9k32743
As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of $(z - z_2k-1)(z - z_2k)$, $k = 1, ldots, frac n2$.
Alternatively, you can show that if $z_1, ldots, z_n$ lie in the same
component of $Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ tag*
g(z)^n = (z-z_1) cdots (z-z_n) , .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^n/2$.
To show the existence of $g$, note that for any closed curve
$gamma$ in $U$ the winding number $N(gamma, z_i)$
is independent of $i$, so that
$$
frac 1n int_gamma left( frac1z-z_1 + ldots + frac1z-z_nright) , dz = N(gamma, z_1)
$$
is a multiple of $2 pi i$. Therefore we can choose $z_0 in U$ and define
$$
g(z) = exp left( frac 1n int_z_0^z left( frac1w-z_1 + ldots + frac1w-z_nright) , dw right) , .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
h(z) = g(z)^-n (z-z_1)cdots(z-z_n)
$$
satisfies
$$
frach'(z)h(z) = -n fracg'(z)g(z) + frac1z-z_1 + ldots + frac1z-z_n = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
answered Jul 15 at 16:18
Martin R
23.9k32743
answered Jul 15 at 16:18
Martin R
23.9k32743
answered Jul 15 at 16:18
Martin R
23.9k32743
answered Jul 15 at 16:18
Martin R
23.9k32743
23.9k32743
add a comment |Â
add a comment |Â
up vote
0
down vote
Assume that $z_2k-1$ lies in the same connected component of $Bbb Csetminus U$ as $z_2k$. Let $w_k(t)$ be a path that connects $w_k(0)=z_2k-1$ with $w_k(1)=z_2k$. Then the homotopy
$$
h(t,z)=prod_k=0^n/2(z-z_2k-1)(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=prod_k=0^n/2(z-z_2k-1)
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
answered Jul 15 at 17:25
LutzL
49.8k31849
add a comment |Â
up vote
0
down vote
Assume that $z_2k-1$ lies in the same connected component of $Bbb Csetminus U$ as $z_2k$. Let $w_k(t)$ be a path that connects $w_k(0)=z_2k-1$ with $w_k(1)=z_2k$. Then the homotopy
$$
h(t,z)=prod_k=0^n/2(z-z_2k-1)(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=prod_k=0^n/2(z-z_2k-1)
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
answered Jul 15 at 17:25
LutzL
49.8k31849
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assume that $z_2k-1$ lies in the same connected component of $Bbb Csetminus U$ as $z_2k$. Let $w_k(t)$ be a path that connects $w_k(0)=z_2k-1$ with $w_k(1)=z_2k$. Then the homotopy
$$
h(t,z)=prod_k=0^n/2(z-z_2k-1)(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=prod_k=0^n/2(z-z_2k-1)
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
answered Jul 15 at 17:25
LutzL
49.8k31849
Assume that $z_2k-1$ lies in the same connected component of $Bbb Csetminus U$ as $z_2k$. Let $w_k(t)$ be a path that connects $w_k(0)=z_2k-1$ with $w_k(1)=z_2k$. Then the homotopy
$$
h(t,z)=prod_k=0^n/2(z-z_2k-1)(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=prod_k=0^n/2(z-z_2k-1)
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
answered Jul 15 at 17:25
LutzL
49.8k31849
answered Jul 15 at 17:25
LutzL
49.8k31849
answered Jul 15 at 17:25
LutzL
49.8k31849
answered Jul 15 at 17:25
LutzL
49.8k31849
49.8k31849
add a comment |Â
add a comment |Â
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2
You can piece the result for $n = 2$ together to get the general case. If $f_k(z)^2 = (z - z_2k-1)(z - z_2k)$, then $f_1cdot f_2 cdot dotsccdot f_n/2$ is a square root of $(z - z_1)cdotdotsccdot(z-z_n)$.
– Daniel Fischer♦
Jul 15 at 13:08
For n=2, I did not need the connectedness of U.Do I need it here?
– Ester
Jul 15 at 13:11
No, connectedness of $U$ just reduces the number of branches to $2$ (if $f$ is one, $-f$ is the other). If $U$ has $m$ connected components, you can choose a sign on each component independently and thus have $2^m$ branches of $sqrt(z-z_1)cdotdotsccdot(z-z_n)$. If $U$ has infinitely many connected components, you have $2^aleph_0$ branches.
– Daniel Fischer♦
Jul 15 at 13:15
So is the result true if U is not assumed to be connected?
– Ester
Jul 15 at 13:17
Yes, it is true whether or not $U$ is connected.
– Daniel Fischer♦
Jul 15 at 13:22