Mixing
holomorphic functions with special condition
$begingroup$
i want to find all holomorphic funtions $ f: mathbb{C} rightarrow mathbb{C}$ with $ f(0)=0$ and $ f(f(f(f(z))))=z forall z in mathbb C$
I would say rotations like $f(z)= |z| e^{i(arg z + frac{pi}{2})} $ can satisfy the conditions?
Am I right?
rotations holomorphic-functions entire-functions
asked Jan 22 at 16:24
Leon1998Leon1998
549
$endgroup$
add a comment |
$begingroup$
i want to find all holomorphic funtions $ f: mathbb{C} rightarrow mathbb{C}$ with $ f(0)=0$ and $ f(f(f(f(z))))=z forall z in mathbb C$
I would say rotations like $f(z)= |z| e^{i(arg z + frac{pi}{2})} $ can satisfy the conditions?
Am I right?
rotations holomorphic-functions entire-functions
asked Jan 22 at 16:24
Leon1998Leon1998
549
$endgroup$
add a comment |
$begingroup$
i want to find all holomorphic funtions $ f: mathbb{C} rightarrow mathbb{C}$ with $ f(0)=0$ and $ f(f(f(f(z))))=z forall z in mathbb C$
I would say rotations like $f(z)= |z| e^{i(arg z + frac{pi}{2})} $ can satisfy the conditions?
Am I right?
rotations holomorphic-functions entire-functions
asked Jan 22 at 16:24
Leon1998Leon1998
549
$endgroup$
i want to find all holomorphic funtions $ f: mathbb{C} rightarrow mathbb{C}$ with $ f(0)=0$ and $ f(f(f(f(z))))=z forall z in mathbb C$
I would say rotations like $f(z)= |z| e^{i(arg z + frac{pi}{2})} $ can satisfy the conditions?
Am I right?
rotations holomorphic-functions entire-functions
rotations holomorphic-functions entire-functions
asked Jan 22 at 16:24
Leon1998Leon1998
549
asked Jan 22 at 16:24
Leon1998Leon1998
549
asked Jan 22 at 16:24
Leon1998Leon1998
549
asked Jan 22 at 16:24
Leon1998Leon1998
549
asked Jan 22 at 16:24
Leon1998Leon1998
549
549
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Such a map is necessarily an automorphism of $mathbb{C}$ and hence of the form $f(z) = az+b$. From here it is easy to see that only rotations qualify. So yes, you already found all of them.
EDIT: Seems like I slightly misread the question. Yes, rotations satisfy this as long as the angles add up to a multiple of $2pi$.
answered Jan 22 at 16:30
KlausKlaus
2,12711
$endgroup$
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
|
show 3 more comments
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Such a map is necessarily an automorphism of $mathbb{C}$ and hence of the form $f(z) = az+b$. From here it is easy to see that only rotations qualify. So yes, you already found all of them.
EDIT: Seems like I slightly misread the question. Yes, rotations satisfy this as long as the angles add up to a multiple of $2pi$.
answered Jan 22 at 16:30
KlausKlaus
2,12711
$endgroup$
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
|
show 3 more comments
$begingroup$
Such a map is necessarily an automorphism of $mathbb{C}$ and hence of the form $f(z) = az+b$. From here it is easy to see that only rotations qualify. So yes, you already found all of them.
EDIT: Seems like I slightly misread the question. Yes, rotations satisfy this as long as the angles add up to a multiple of $2pi$.
answered Jan 22 at 16:30
KlausKlaus
2,12711
$endgroup$
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
|
show 3 more comments
$begingroup$
Such a map is necessarily an automorphism of $mathbb{C}$ and hence of the form $f(z) = az+b$. From here it is easy to see that only rotations qualify. So yes, you already found all of them.
EDIT: Seems like I slightly misread the question. Yes, rotations satisfy this as long as the angles add up to a multiple of $2pi$.
answered Jan 22 at 16:30
KlausKlaus
2,12711
$endgroup$
Such a map is necessarily an automorphism of $mathbb{C}$ and hence of the form $f(z) = az+b$. From here it is easy to see that only rotations qualify. So yes, you already found all of them.
EDIT: Seems like I slightly misread the question. Yes, rotations satisfy this as long as the angles add up to a multiple of $2pi$.
answered Jan 22 at 16:30
KlausKlaus
2,12711
answered Jan 22 at 16:30
KlausKlaus
2,12711
answered Jan 22 at 16:30
KlausKlaus
2,12711
answered Jan 22 at 16:30
KlausKlaus
2,12711
2,12711
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
|
show 3 more comments
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How can i adapt my function then?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
How do you conlude from $f(z)=az +b$ to rotations?
$endgroup$
– Leon1998
Jan 22 at 16:36
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
I do not understand your first comment. Why do you want to adapt it? As for your second comment, $b$ has to be $0$ because you demanded $f(0) = 0$. Hence $f(z) = az$ and $f(f(f(f(z)))) = a^4z$, i.e. $a in {1,i,-1,-i}$. Therefore $f$ is a rotation. Your example is in fact $f(z) = iz$.
$endgroup$
– Klaus
Jan 22 at 16:45
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
But I thought I have to rotate 90 degrees each, so 4 times? In your function $f(z)=iz$ you don*t consider that? I'm sure that I missed something:(
$endgroup$
– Leon1998
Jan 22 at 16:52
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
$begingroup$
$f(z) = iz$ $is$ the rotation about 90 degrees.
$endgroup$
– Klaus
Jan 22 at 16:54
|
show 3 more comments
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