Mixing
Branches of $frac{1}{1-cos (zsqrt z)}$
$begingroup$
I was wondering why should $f(z)=frac{1}{1-cos (zsqrt z)}$ be a monodromic function. By the properties of the complex square root one has:
$z sqrt z=z z^{1/2}=z^{3/2}:=e^{3/2 ln_{mathbb{C}}(z)}=|z|^{3/2}e^{3/2 iarg z}e^{3 k pi i}$, where $k in mathbb{Z}$.
If $|k|$ is even: $z sqrt z = |z|^{3/2}e^{3/2 iarg z}$, otherwise $z sqrt z =-|z|^{3/2}e^{3/2 iarg z}$. So I've ended up with two branches and $f$ is not monodromic.
Have I committed a mistake?
Thanks!
complex-analysis
asked Jan 14 at 17:56
LeonardoLeonardo
3449
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add a comment |
$begingroup$
I was wondering why should $f(z)=frac{1}{1-cos (zsqrt z)}$ be a monodromic function. By the properties of the complex square root one has:
$z sqrt z=z z^{1/2}=z^{3/2}:=e^{3/2 ln_{mathbb{C}}(z)}=|z|^{3/2}e^{3/2 iarg z}e^{3 k pi i}$, where $k in mathbb{Z}$.
If $|k|$ is even: $z sqrt z = |z|^{3/2}e^{3/2 iarg z}$, otherwise $z sqrt z =-|z|^{3/2}e^{3/2 iarg z}$. So I've ended up with two branches and $f$ is not monodromic.
Have I committed a mistake?
Thanks!
complex-analysis
asked Jan 14 at 17:56
LeonardoLeonardo
3449
$endgroup$
add a comment |
$begingroup$
I was wondering why should $f(z)=frac{1}{1-cos (zsqrt z)}$ be a monodromic function. By the properties of the complex square root one has:
$z sqrt z=z z^{1/2}=z^{3/2}:=e^{3/2 ln_{mathbb{C}}(z)}=|z|^{3/2}e^{3/2 iarg z}e^{3 k pi i}$, where $k in mathbb{Z}$.
If $|k|$ is even: $z sqrt z = |z|^{3/2}e^{3/2 iarg z}$, otherwise $z sqrt z =-|z|^{3/2}e^{3/2 iarg z}$. So I've ended up with two branches and $f$ is not monodromic.
Have I committed a mistake?
Thanks!
complex-analysis
asked Jan 14 at 17:56
LeonardoLeonardo
3449
$endgroup$
I was wondering why should $f(z)=frac{1}{1-cos (zsqrt z)}$ be a monodromic function. By the properties of the complex square root one has:
$z sqrt z=z z^{1/2}=z^{3/2}:=e^{3/2 ln_{mathbb{C}}(z)}=|z|^{3/2}e^{3/2 iarg z}e^{3 k pi i}$, where $k in mathbb{Z}$.
If $|k|$ is even: $z sqrt z = |z|^{3/2}e^{3/2 iarg z}$, otherwise $z sqrt z =-|z|^{3/2}e^{3/2 iarg z}$. So I've ended up with two branches and $f$ is not monodromic.
Have I committed a mistake?
Thanks!
complex-analysis
complex-analysis
asked Jan 14 at 17:56
LeonardoLeonardo
3449
asked Jan 14 at 17:56
LeonardoLeonardo
3449
edited Jan 14 at 18:02
Leonardo
asked Jan 14 at 17:56
LeonardoLeonardo
3449
asked Jan 14 at 17:56
LeonardoLeonardo
3449
asked Jan 14 at 17:56
LeonardoLeonardo
3449
3449
add a comment |
add a comment |
1 Answer
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The key is that $cos$ is an even function. Its power series at zero has only even-degree terms, so if we substitute $zsqrt{z}$ there, we get a new series with only degrees divisible by $3$. Therefore $cos(zsqrt{z})$ is an entire function, and $frac1{1-cos(zsqrt{z})}$ is meromorphic without branches on the whole plane.
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
$endgroup$
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The key is that $cos$ is an even function. Its power series at zero has only even-degree terms, so if we substitute $zsqrt{z}$ there, we get a new series with only degrees divisible by $3$. Therefore $cos(zsqrt{z})$ is an entire function, and $frac1{1-cos(zsqrt{z})}$ is meromorphic without branches on the whole plane.
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
$endgroup$
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
add a comment |
$begingroup$
The key is that $cos$ is an even function. Its power series at zero has only even-degree terms, so if we substitute $zsqrt{z}$ there, we get a new series with only degrees divisible by $3$. Therefore $cos(zsqrt{z})$ is an entire function, and $frac1{1-cos(zsqrt{z})}$ is meromorphic without branches on the whole plane.
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
$endgroup$
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
add a comment |
$begingroup$
The key is that $cos$ is an even function. Its power series at zero has only even-degree terms, so if we substitute $zsqrt{z}$ there, we get a new series with only degrees divisible by $3$. Therefore $cos(zsqrt{z})$ is an entire function, and $frac1{1-cos(zsqrt{z})}$ is meromorphic without branches on the whole plane.
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
$endgroup$
The key is that $cos$ is an even function. Its power series at zero has only even-degree terms, so if we substitute $zsqrt{z}$ there, we get a new series with only degrees divisible by $3$. Therefore $cos(zsqrt{z})$ is an entire function, and $frac1{1-cos(zsqrt{z})}$ is meromorphic without branches on the whole plane.
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
answered Jan 14 at 18:17
jmerryjmerry
9,1271123
9,1271123
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
add a comment |
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
1
1
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
$begingroup$
(+1) This is a point not often discussed in a course in complex analysis, but is very useful to the applied mathematician, physicist, engineer, etc. And without even discussing the power series of the cosine we see that $cos(f(z))=frac12(e^{if(z)}+e^{-if(z)})$ is even in $f(z)$.
$endgroup$
– Mark Viola
Jan 14 at 18:27
add a comment |
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