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Complex function does not have a modulus equal to a function of its real part
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An analytic function $f(z) = f(x+iy)$ in $mathbbC$ cannot have modulus $fracAcosh(x)$ for some constant $A neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (fracAcosh x)^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.
My start point is:
$$u_x = v_y text and u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 fracsinh x(cosh x )^2$$
and
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!
complex-analysis analysis functions trigonometry
asked 2 hours ago
rannoudanames
406614
add a comment |Â
up vote
2
down vote
favorite
An analytic function $f(z) = f(x+iy)$ in $mathbbC$ cannot have modulus $fracAcosh(x)$ for some constant $A neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (fracAcosh x)^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.
My start point is:
$$u_x = v_y text and u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 fracsinh x(cosh x )^2$$
and
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!
complex-analysis analysis functions trigonometry
asked 2 hours ago
rannoudanames
406614
1
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
1
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
An analytic function $f(z) = f(x+iy)$ in $mathbbC$ cannot have modulus $fracAcosh(x)$ for some constant $A neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (fracAcosh x)^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.
My start point is:
$$u_x = v_y text and u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 fracsinh x(cosh x )^2$$
and
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!
complex-analysis analysis functions trigonometry
asked 2 hours ago
rannoudanames
406614
An analytic function $f(z) = f(x+iy)$ in $mathbbC$ cannot have modulus $fracAcosh(x)$ for some constant $A neq 0$.
Can we do so simply using the Cauchy-Riemann Equations?
I tried working by contradiction:
Say there is such a $f(z)$. Given $f(z) = u(x,y) + i v(x,y)$ is analytic, it satisfies: $u_x = v_y$ and $u_y = -v_x$
We also see that: $$|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (fracAcosh x)^2$$
I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations.
My start point is:
$$u_x = v_y text and u_y = -v_x$$
along with
$$u(x,y)u_x + v(x,y)v_x = -A^2 fracsinh x(cosh x )^2$$
and
$$u(x,y)u_y + v(x,y)v_y = 0$$
I would appreciate some hints!
complex-analysis analysis functions trigonometry
complex-analysis analysis functions trigonometry
asked 2 hours ago
rannoudanames
406614
asked 2 hours ago
rannoudanames
406614
edited 1 hour ago
asked 2 hours ago
rannoudanames
406614
asked 2 hours ago
rannoudanames
406614
asked 2 hours ago
rannoudanames
406614
406614
1
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
1
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago
add a comment |Â
1
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
1
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago
1
1
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
1
1
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
So
$$
beginalign*
uu_x+vv_x&=-A^2fracsinh xcosh^3 x\
-uv_x+vu_x&=0\
endalign*
$$
So
$$
u_x=-ufracsinh xcosh x
$$
(remember $u^2+v^2=dfracA^2cosh^2 x$) and similarly $v_x=-vdfracsinh xcosh x$. So
$$
u=frac1cosh x+f(y)text and v=frac1cosh x+g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
thank you very much !!
– rannoudanames
1 hour ago
add a comment |Â
up vote
2
down vote
We could look at it this way:
If $f(z) ne 0$ is holomorphic, then $ln vert f(z) vert$ is harmonic.
For if $f(z) = u(z) + iv(z) ne 0$ is holomorphic, locally we may write
$f(z) = r(z) e^itheta(z), tag 1$
where
$r(z) = vert f(z) vert ne 0, tag 2$
and
$theta(z) = arg (f(z)) = tan^-1 dfracv(z)u(z) tag 3$
when $u(z) ne 0$, and
$theta(z) = arg (f(z)) = cot^-1 dfracu(z)v(z) tag 4$
when $v(z) ne 0$; note $u(z)$ and $v(z)$ cannot both be zero since $f(z) ne 0$.
Now since $f(z) ne 0$ is holomorphic, so is $ln f(z)$; we have from (1)
$ln f(z) = ln r(z) + itheta(z); tag 5$
it follows that
$ln r(z) = ln vert f(z) vert tag 6$
is harmonic, being the real part of the holomorphic function $ln f(z)$.
Now if
$vert f(z) vert = A(cosh x)^-1, tag 7$
the from what we have done above, $ln A(cosh x)^-1$ is harmonic; but it is easily computed that
$nabla^2 ln (A(cosh x)^-1) = left ( dfracpartial^2partial x^2 + dfracpartial^2partial y^2 right ) ln (A(cosh x)^-1) = dfracpartial^2partial x^2 ln (A(cosh x)^-1) ne 0; tag 8$
it follows that $ln A(cosh x)^-1$ is not harmonic, hence $A(cosh x)^-1$is not the modulus of any holomorphic function $f(z)$.
Note: If $f(z)$ is entire, we can also argue from Liouville's theorem:
$vert f(z) vert = A(cosh x)^-1$ is bounded; but a bounded entire function is constant, so $vert f(z) vert = A(cosh x)^-1$ is impossible.
answered 1 hour ago
Robert Lewis
38.6k22358
add a comment |Â
up vote
0
down vote
Here is another (perhaps interesting) proof: let $g(z)=(e^z+e^-z)f(z)$. Then $|g(z)|leq 2A $ because $|e^z+e^-z| leq e^x+e^-x =2cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c neq 0$. We have $(e^z+e^-z)f(z)=c$. You get a contradiction by taking $z=ipi /2$
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
So
$$
beginalign*
uu_x+vv_x&=-A^2fracsinh xcosh^3 x\
-uv_x+vu_x&=0\
endalign*
$$
So
$$
u_x=-ufracsinh xcosh x
$$
(remember $u^2+v^2=dfracA^2cosh^2 x$) and similarly $v_x=-vdfracsinh xcosh x$. So
$$
u=frac1cosh x+f(y)text and v=frac1cosh x+g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
thank you very much !!
– rannoudanames
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
So
$$
beginalign*
uu_x+vv_x&=-A^2fracsinh xcosh^3 x\
-uv_x+vu_x&=0\
endalign*
$$
So
$$
u_x=-ufracsinh xcosh x
$$
(remember $u^2+v^2=dfracA^2cosh^2 x$) and similarly $v_x=-vdfracsinh xcosh x$. So
$$
u=frac1cosh x+f(y)text and v=frac1cosh x+g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
thank you very much !!
– rannoudanames
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
So
$$
beginalign*
uu_x+vv_x&=-A^2fracsinh xcosh^3 x\
-uv_x+vu_x&=0\
endalign*
$$
So
$$
u_x=-ufracsinh xcosh x
$$
(remember $u^2+v^2=dfracA^2cosh^2 x$) and similarly $v_x=-vdfracsinh xcosh x$. So
$$
u=frac1cosh x+f(y)text and v=frac1cosh x+g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So
$$
beginalign*
uu_x+vv_x&=-A^2fracsinh xcosh^3 x\
-uv_x+vu_x&=0\
endalign*
$$
So
$$
u_x=-ufracsinh xcosh x
$$
(remember $u^2+v^2=dfracA^2cosh^2 x$) and similarly $v_x=-vdfracsinh xcosh x$. So
$$
u=frac1cosh x+f(y)text and v=frac1cosh x+g(y)
$$
and there are no $g$ that gives $u_x=v_y$.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
user10354138
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
user10354138
8645
answered 1 hour ago
user10354138
8645
8645
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
thank you very much !!
– rannoudanames
1 hour ago
add a comment |Â
1
thank you very much !!
– rannoudanames
1 hour ago
1
1
thank you very much !!
– rannoudanames
1 hour ago
thank you very much !!
– rannoudanames
1 hour ago
add a comment |Â
up vote
2
down vote
We could look at it this way:
If $f(z) ne 0$ is holomorphic, then $ln vert f(z) vert$ is harmonic.
For if $f(z) = u(z) + iv(z) ne 0$ is holomorphic, locally we may write
$f(z) = r(z) e^itheta(z), tag 1$
where
$r(z) = vert f(z) vert ne 0, tag 2$
and
$theta(z) = arg (f(z)) = tan^-1 dfracv(z)u(z) tag 3$
when $u(z) ne 0$, and
$theta(z) = arg (f(z)) = cot^-1 dfracu(z)v(z) tag 4$
when $v(z) ne 0$; note $u(z)$ and $v(z)$ cannot both be zero since $f(z) ne 0$.
Now since $f(z) ne 0$ is holomorphic, so is $ln f(z)$; we have from (1)
$ln f(z) = ln r(z) + itheta(z); tag 5$
it follows that
$ln r(z) = ln vert f(z) vert tag 6$
is harmonic, being the real part of the holomorphic function $ln f(z)$.
Now if
$vert f(z) vert = A(cosh x)^-1, tag 7$
the from what we have done above, $ln A(cosh x)^-1$ is harmonic; but it is easily computed that
$nabla^2 ln (A(cosh x)^-1) = left ( dfracpartial^2partial x^2 + dfracpartial^2partial y^2 right ) ln (A(cosh x)^-1) = dfracpartial^2partial x^2 ln (A(cosh x)^-1) ne 0; tag 8$
it follows that $ln A(cosh x)^-1$ is not harmonic, hence $A(cosh x)^-1$is not the modulus of any holomorphic function $f(z)$.
Note: If $f(z)$ is entire, we can also argue from Liouville's theorem:
$vert f(z) vert = A(cosh x)^-1$ is bounded; but a bounded entire function is constant, so $vert f(z) vert = A(cosh x)^-1$ is impossible.
answered 1 hour ago
Robert Lewis
38.6k22358
add a comment |Â
up vote
2
down vote
We could look at it this way:
If $f(z) ne 0$ is holomorphic, then $ln vert f(z) vert$ is harmonic.
For if $f(z) = u(z) + iv(z) ne 0$ is holomorphic, locally we may write
$f(z) = r(z) e^itheta(z), tag 1$
where
$r(z) = vert f(z) vert ne 0, tag 2$
and
$theta(z) = arg (f(z)) = tan^-1 dfracv(z)u(z) tag 3$
when $u(z) ne 0$, and
$theta(z) = arg (f(z)) = cot^-1 dfracu(z)v(z) tag 4$
when $v(z) ne 0$; note $u(z)$ and $v(z)$ cannot both be zero since $f(z) ne 0$.
Now since $f(z) ne 0$ is holomorphic, so is $ln f(z)$; we have from (1)
$ln f(z) = ln r(z) + itheta(z); tag 5$
it follows that
$ln r(z) = ln vert f(z) vert tag 6$
is harmonic, being the real part of the holomorphic function $ln f(z)$.
Now if
$vert f(z) vert = A(cosh x)^-1, tag 7$
the from what we have done above, $ln A(cosh x)^-1$ is harmonic; but it is easily computed that
$nabla^2 ln (A(cosh x)^-1) = left ( dfracpartial^2partial x^2 + dfracpartial^2partial y^2 right ) ln (A(cosh x)^-1) = dfracpartial^2partial x^2 ln (A(cosh x)^-1) ne 0; tag 8$
it follows that $ln A(cosh x)^-1$ is not harmonic, hence $A(cosh x)^-1$is not the modulus of any holomorphic function $f(z)$.
Note: If $f(z)$ is entire, we can also argue from Liouville's theorem:
$vert f(z) vert = A(cosh x)^-1$ is bounded; but a bounded entire function is constant, so $vert f(z) vert = A(cosh x)^-1$ is impossible.
answered 1 hour ago
Robert Lewis
38.6k22358
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We could look at it this way:
If $f(z) ne 0$ is holomorphic, then $ln vert f(z) vert$ is harmonic.
For if $f(z) = u(z) + iv(z) ne 0$ is holomorphic, locally we may write
$f(z) = r(z) e^itheta(z), tag 1$
where
$r(z) = vert f(z) vert ne 0, tag 2$
and
$theta(z) = arg (f(z)) = tan^-1 dfracv(z)u(z) tag 3$
when $u(z) ne 0$, and
$theta(z) = arg (f(z)) = cot^-1 dfracu(z)v(z) tag 4$
when $v(z) ne 0$; note $u(z)$ and $v(z)$ cannot both be zero since $f(z) ne 0$.
Now since $f(z) ne 0$ is holomorphic, so is $ln f(z)$; we have from (1)
$ln f(z) = ln r(z) + itheta(z); tag 5$
it follows that
$ln r(z) = ln vert f(z) vert tag 6$
is harmonic, being the real part of the holomorphic function $ln f(z)$.
Now if
$vert f(z) vert = A(cosh x)^-1, tag 7$
the from what we have done above, $ln A(cosh x)^-1$ is harmonic; but it is easily computed that
$nabla^2 ln (A(cosh x)^-1) = left ( dfracpartial^2partial x^2 + dfracpartial^2partial y^2 right ) ln (A(cosh x)^-1) = dfracpartial^2partial x^2 ln (A(cosh x)^-1) ne 0; tag 8$
it follows that $ln A(cosh x)^-1$ is not harmonic, hence $A(cosh x)^-1$is not the modulus of any holomorphic function $f(z)$.
Note: If $f(z)$ is entire, we can also argue from Liouville's theorem:
$vert f(z) vert = A(cosh x)^-1$ is bounded; but a bounded entire function is constant, so $vert f(z) vert = A(cosh x)^-1$ is impossible.
answered 1 hour ago
Robert Lewis
38.6k22358
We could look at it this way:
If $f(z) ne 0$ is holomorphic, then $ln vert f(z) vert$ is harmonic.
For if $f(z) = u(z) + iv(z) ne 0$ is holomorphic, locally we may write
$f(z) = r(z) e^itheta(z), tag 1$
where
$r(z) = vert f(z) vert ne 0, tag 2$
and
$theta(z) = arg (f(z)) = tan^-1 dfracv(z)u(z) tag 3$
when $u(z) ne 0$, and
$theta(z) = arg (f(z)) = cot^-1 dfracu(z)v(z) tag 4$
when $v(z) ne 0$; note $u(z)$ and $v(z)$ cannot both be zero since $f(z) ne 0$.
Now since $f(z) ne 0$ is holomorphic, so is $ln f(z)$; we have from (1)
$ln f(z) = ln r(z) + itheta(z); tag 5$
it follows that
$ln r(z) = ln vert f(z) vert tag 6$
is harmonic, being the real part of the holomorphic function $ln f(z)$.
Now if
$vert f(z) vert = A(cosh x)^-1, tag 7$
the from what we have done above, $ln A(cosh x)^-1$ is harmonic; but it is easily computed that
$nabla^2 ln (A(cosh x)^-1) = left ( dfracpartial^2partial x^2 + dfracpartial^2partial y^2 right ) ln (A(cosh x)^-1) = dfracpartial^2partial x^2 ln (A(cosh x)^-1) ne 0; tag 8$
it follows that $ln A(cosh x)^-1$ is not harmonic, hence $A(cosh x)^-1$is not the modulus of any holomorphic function $f(z)$.
Note: If $f(z)$ is entire, we can also argue from Liouville's theorem:
$vert f(z) vert = A(cosh x)^-1$ is bounded; but a bounded entire function is constant, so $vert f(z) vert = A(cosh x)^-1$ is impossible.
answered 1 hour ago
Robert Lewis
38.6k22358
edited 32 mins ago
answered 1 hour ago
Robert Lewis
38.6k22358
answered 1 hour ago
Robert Lewis
38.6k22358
answered 1 hour ago
Robert Lewis
38.6k22358
38.6k22358
add a comment |Â
add a comment |Â
up vote
0
down vote
Here is another (perhaps interesting) proof: let $g(z)=(e^z+e^-z)f(z)$. Then $|g(z)|leq 2A $ because $|e^z+e^-z| leq e^x+e^-x =2cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c neq 0$. We have $(e^z+e^-z)f(z)=c$. You get a contradiction by taking $z=ipi /2$
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
add a comment |Â
up vote
0
down vote
Here is another (perhaps interesting) proof: let $g(z)=(e^z+e^-z)f(z)$. Then $|g(z)|leq 2A $ because $|e^z+e^-z| leq e^x+e^-x =2cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c neq 0$. We have $(e^z+e^-z)f(z)=c$. You get a contradiction by taking $z=ipi /2$
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is another (perhaps interesting) proof: let $g(z)=(e^z+e^-z)f(z)$. Then $|g(z)|leq 2A $ because $|e^z+e^-z| leq e^x+e^-x =2cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c neq 0$. We have $(e^z+e^-z)f(z)=c$. You get a contradiction by taking $z=ipi /2$
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
Here is another (perhaps interesting) proof: let $g(z)=(e^z+e^-z)f(z)$. Then $|g(z)|leq 2A $ because $|e^z+e^-z| leq e^x+e^-x =2cosh x$. By Louiville's Theorem $g$ is a constant, say $c$. Clearly, $c neq 0$. We have $(e^z+e^-z)f(z)=c$. You get a contradiction by taking $z=ipi /2$
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
answered 14 mins ago
Kavi Rama Murthy
26.1k31437
26.1k31437
add a comment |Â
add a comment |Â
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1
Please do not delete and repost the same question. To add to a question, edit the original post, instead.
– dxiv
2 hours ago
1
@dxiv Sorry! I was not aware! Will do next time!
– rannoudanames
2 hours ago