Analytic function must be free from $bar{z}.$
How can i say that analytic function must be free from $bar{z}?$ I verified it for many examples like $bar{z}, z^2+bar{z} $ etc and found that the results seems to be true according to me . I am thinking like as if any complex function contains $bar{z}$ term in any form(obviously in nontrivial form unlike $bar{z}-bar{z}$) then that terms is never analytic and hence whole function is also never analytic. Please suggest . Thanks .
complex-analysis
add a comment |
How can i say that analytic function must be free from $bar{z}?$ I verified it for many examples like $bar{z}, z^2+bar{z} $ etc and found that the results seems to be true according to me . I am thinking like as if any complex function contains $bar{z}$ term in any form(obviously in nontrivial form unlike $bar{z}-bar{z}$) then that terms is never analytic and hence whole function is also never analytic. Please suggest . Thanks .
complex-analysis
$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53
add a comment |
How can i say that analytic function must be free from $bar{z}?$ I verified it for many examples like $bar{z}, z^2+bar{z} $ etc and found that the results seems to be true according to me . I am thinking like as if any complex function contains $bar{z}$ term in any form(obviously in nontrivial form unlike $bar{z}-bar{z}$) then that terms is never analytic and hence whole function is also never analytic. Please suggest . Thanks .
complex-analysis
How can i say that analytic function must be free from $bar{z}?$ I verified it for many examples like $bar{z}, z^2+bar{z} $ etc and found that the results seems to be true according to me . I am thinking like as if any complex function contains $bar{z}$ term in any form(obviously in nontrivial form unlike $bar{z}-bar{z}$) then that terms is never analytic and hence whole function is also never analytic. Please suggest . Thanks .
complex-analysis
complex-analysis
asked Nov 23 at 6:46
neelkanth
2,0541927
2,0541927
$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53
add a comment |
$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53
$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53
add a comment |
1 Answer
1
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This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={rm conjugate}(bar z)$ is analytic $ldots$
I'm going to argue in the following setup: Let
$$Phi:quadOmegato{mathbb C},qquad(z,zeta)toPhi(z,zeta)$$
be defined on the domain $Omegasubset{mathbb C}^2$. Assume that $Phi$ is (i) a $C^1$ function of its four real variables $(x, y, xi,eta)$, and is (ii) separately analytic in each of the two complex variables $z$ and $zeta$, meaning that the partial functions
$$zmapstoPhi(z,zeta_0),qquad {rm resp.,}qquad zetamapstoPhi(z_0,zeta)$$
are analytic functions of $z$, resp., of $zeta$, as long as $(z,zeta_0)inOmega$, resp., $(z_0,zeta)inOmega$. Such a $Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $zeta$.
Given such a $Phi$ we consider the function
$$f(z):=Phi(z,bar z)qquad(zinOmega') ,tag{1}$$
whose domain is the open set $Omega':=bigl{zin{mathbb C}bigm| (z,bar z)inOmegabigr}subset{mathbb C}$. We now pose the following question:
What are necessary and sufficient conditions on $Phi$ making $f$ an analytic function on $Omega'$?
Claim. The function $f$ is analytic on $Omega'$ iff $$Phi_{.2}(z,bar z)=0qquadforall>zinOmega' .$$
Here $Phi_{.2}$ denotes the partial derivative of $Phi$ with respect to its second complex variable.
Proof. Fix a $zinOmega'$, and let $h$ be a complex increment variable. Then
$$eqalign{f(z+h)-f(z)&=Phi(z+h,bar z+bar h)-Phi(z,bar z)cr &=Phi_{.1}(z,bar z)>h+Phi_{.2}(z,bar z)>bar h +obigl(|h|bigr)qquad(hto0in{mathbb C}) .cr}tag{2}$$
Here we just have used the assumptions on $Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $hmapsto lambda h$ for some $lambdain{mathbb C}$, iff $Phi_{.2}(z,bar z)=0$.$quadsquare$
When we are applying this principle there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly introduce an auxiliary variable $zeta$. Instead one just considers the $bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $Phi(z,bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${partialPhi(z,bar z)overpartialbar z}=0qquadforall zinOmega' .$$
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
add a comment |
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This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={rm conjugate}(bar z)$ is analytic $ldots$
I'm going to argue in the following setup: Let
$$Phi:quadOmegato{mathbb C},qquad(z,zeta)toPhi(z,zeta)$$
be defined on the domain $Omegasubset{mathbb C}^2$. Assume that $Phi$ is (i) a $C^1$ function of its four real variables $(x, y, xi,eta)$, and is (ii) separately analytic in each of the two complex variables $z$ and $zeta$, meaning that the partial functions
$$zmapstoPhi(z,zeta_0),qquad {rm resp.,}qquad zetamapstoPhi(z_0,zeta)$$
are analytic functions of $z$, resp., of $zeta$, as long as $(z,zeta_0)inOmega$, resp., $(z_0,zeta)inOmega$. Such a $Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $zeta$.
Given such a $Phi$ we consider the function
$$f(z):=Phi(z,bar z)qquad(zinOmega') ,tag{1}$$
whose domain is the open set $Omega':=bigl{zin{mathbb C}bigm| (z,bar z)inOmegabigr}subset{mathbb C}$. We now pose the following question:
What are necessary and sufficient conditions on $Phi$ making $f$ an analytic function on $Omega'$?
Claim. The function $f$ is analytic on $Omega'$ iff $$Phi_{.2}(z,bar z)=0qquadforall>zinOmega' .$$
Here $Phi_{.2}$ denotes the partial derivative of $Phi$ with respect to its second complex variable.
Proof. Fix a $zinOmega'$, and let $h$ be a complex increment variable. Then
$$eqalign{f(z+h)-f(z)&=Phi(z+h,bar z+bar h)-Phi(z,bar z)cr &=Phi_{.1}(z,bar z)>h+Phi_{.2}(z,bar z)>bar h +obigl(|h|bigr)qquad(hto0in{mathbb C}) .cr}tag{2}$$
Here we just have used the assumptions on $Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $hmapsto lambda h$ for some $lambdain{mathbb C}$, iff $Phi_{.2}(z,bar z)=0$.$quadsquare$
When we are applying this principle there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly introduce an auxiliary variable $zeta$. Instead one just considers the $bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $Phi(z,bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${partialPhi(z,bar z)overpartialbar z}=0qquadforall zinOmega' .$$
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
add a comment |
This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={rm conjugate}(bar z)$ is analytic $ldots$
I'm going to argue in the following setup: Let
$$Phi:quadOmegato{mathbb C},qquad(z,zeta)toPhi(z,zeta)$$
be defined on the domain $Omegasubset{mathbb C}^2$. Assume that $Phi$ is (i) a $C^1$ function of its four real variables $(x, y, xi,eta)$, and is (ii) separately analytic in each of the two complex variables $z$ and $zeta$, meaning that the partial functions
$$zmapstoPhi(z,zeta_0),qquad {rm resp.,}qquad zetamapstoPhi(z_0,zeta)$$
are analytic functions of $z$, resp., of $zeta$, as long as $(z,zeta_0)inOmega$, resp., $(z_0,zeta)inOmega$. Such a $Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $zeta$.
Given such a $Phi$ we consider the function
$$f(z):=Phi(z,bar z)qquad(zinOmega') ,tag{1}$$
whose domain is the open set $Omega':=bigl{zin{mathbb C}bigm| (z,bar z)inOmegabigr}subset{mathbb C}$. We now pose the following question:
What are necessary and sufficient conditions on $Phi$ making $f$ an analytic function on $Omega'$?
Claim. The function $f$ is analytic on $Omega'$ iff $$Phi_{.2}(z,bar z)=0qquadforall>zinOmega' .$$
Here $Phi_{.2}$ denotes the partial derivative of $Phi$ with respect to its second complex variable.
Proof. Fix a $zinOmega'$, and let $h$ be a complex increment variable. Then
$$eqalign{f(z+h)-f(z)&=Phi(z+h,bar z+bar h)-Phi(z,bar z)cr &=Phi_{.1}(z,bar z)>h+Phi_{.2}(z,bar z)>bar h +obigl(|h|bigr)qquad(hto0in{mathbb C}) .cr}tag{2}$$
Here we just have used the assumptions on $Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $hmapsto lambda h$ for some $lambdain{mathbb C}$, iff $Phi_{.2}(z,bar z)=0$.$quadsquare$
When we are applying this principle there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly introduce an auxiliary variable $zeta$. Instead one just considers the $bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $Phi(z,bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${partialPhi(z,bar z)overpartialbar z}=0qquadforall zinOmega' .$$
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
add a comment |
This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={rm conjugate}(bar z)$ is analytic $ldots$
I'm going to argue in the following setup: Let
$$Phi:quadOmegato{mathbb C},qquad(z,zeta)toPhi(z,zeta)$$
be defined on the domain $Omegasubset{mathbb C}^2$. Assume that $Phi$ is (i) a $C^1$ function of its four real variables $(x, y, xi,eta)$, and is (ii) separately analytic in each of the two complex variables $z$ and $zeta$, meaning that the partial functions
$$zmapstoPhi(z,zeta_0),qquad {rm resp.,}qquad zetamapstoPhi(z_0,zeta)$$
are analytic functions of $z$, resp., of $zeta$, as long as $(z,zeta_0)inOmega$, resp., $(z_0,zeta)inOmega$. Such a $Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $zeta$.
Given such a $Phi$ we consider the function
$$f(z):=Phi(z,bar z)qquad(zinOmega') ,tag{1}$$
whose domain is the open set $Omega':=bigl{zin{mathbb C}bigm| (z,bar z)inOmegabigr}subset{mathbb C}$. We now pose the following question:
What are necessary and sufficient conditions on $Phi$ making $f$ an analytic function on $Omega'$?
Claim. The function $f$ is analytic on $Omega'$ iff $$Phi_{.2}(z,bar z)=0qquadforall>zinOmega' .$$
Here $Phi_{.2}$ denotes the partial derivative of $Phi$ with respect to its second complex variable.
Proof. Fix a $zinOmega'$, and let $h$ be a complex increment variable. Then
$$eqalign{f(z+h)-f(z)&=Phi(z+h,bar z+bar h)-Phi(z,bar z)cr &=Phi_{.1}(z,bar z)>h+Phi_{.2}(z,bar z)>bar h +obigl(|h|bigr)qquad(hto0in{mathbb C}) .cr}tag{2}$$
Here we just have used the assumptions on $Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $hmapsto lambda h$ for some $lambdain{mathbb C}$, iff $Phi_{.2}(z,bar z)=0$.$quadsquare$
When we are applying this principle there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly introduce an auxiliary variable $zeta$. Instead one just considers the $bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $Phi(z,bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${partialPhi(z,bar z)overpartialbar z}=0qquadforall zinOmega' .$$
This is an interesting question, which is usually glossed over when doing complex calculus in $z$ and $bar z$ terms. It is also not so clear what the actual claim is. E.g., the function $q(z):={rm conjugate}(bar z)$ is analytic $ldots$
I'm going to argue in the following setup: Let
$$Phi:quadOmegato{mathbb C},qquad(z,zeta)toPhi(z,zeta)$$
be defined on the domain $Omegasubset{mathbb C}^2$. Assume that $Phi$ is (i) a $C^1$ function of its four real variables $(x, y, xi,eta)$, and is (ii) separately analytic in each of the two complex variables $z$ and $zeta$, meaning that the partial functions
$$zmapstoPhi(z,zeta_0),qquad {rm resp.,}qquad zetamapstoPhi(z_0,zeta)$$
are analytic functions of $z$, resp., of $zeta$, as long as $(z,zeta_0)inOmega$, resp., $(z_0,zeta)inOmega$. Such a $Phi$ could be given as a series, or as an "analytic expression" in terms of the variables $z$ and $zeta$.
Given such a $Phi$ we consider the function
$$f(z):=Phi(z,bar z)qquad(zinOmega') ,tag{1}$$
whose domain is the open set $Omega':=bigl{zin{mathbb C}bigm| (z,bar z)inOmegabigr}subset{mathbb C}$. We now pose the following question:
What are necessary and sufficient conditions on $Phi$ making $f$ an analytic function on $Omega'$?
Claim. The function $f$ is analytic on $Omega'$ iff $$Phi_{.2}(z,bar z)=0qquadforall>zinOmega' .$$
Here $Phi_{.2}$ denotes the partial derivative of $Phi$ with respect to its second complex variable.
Proof. Fix a $zinOmega'$, and let $h$ be a complex increment variable. Then
$$eqalign{f(z+h)-f(z)&=Phi(z+h,bar z+bar h)-Phi(z,bar z)cr &=Phi_{.1}(z,bar z)>h+Phi_{.2}(z,bar z)>bar h +obigl(|h|bigr)qquad(hto0in{mathbb C}) .cr}tag{2}$$
Here we just have used the assumptions on $Phi$ and the definition of derivative. Now the main part of the right hand side of $(2)$ is a complex linear function of $h$, i.e., of the form $hmapsto lambda h$ for some $lambdain{mathbb C}$, iff $Phi_{.2}(z,bar z)=0$.$quadsquare$
When we are applying this principle there will be given a function $f$ defined in the form $(1)$. By abuse of notation one then does not explicitly introduce an auxiliary variable $zeta$. Instead one just considers the $bar z$ appearing in $(1)$ as an "independent complex variable" and formally differentiates the expression $Phi(z,bar z)$ with respect to this variable. In this way one arrives at the handy condition
$${partialPhi(z,bar z)overpartialbar z}=0qquadforall zinOmega' .$$
edited Nov 24 at 17:10
answered Nov 23 at 19:18
Christian Blatter
172k7112325
172k7112325
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
add a comment |
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
"which is usually glossed over when doing complex calculus in z and z¯ terms. It is also not so clear what the actual claim is." No. The approach you suggest is ingenious but, to "sell" it, it is not necessary to cast aspersions on the way complex analysis is taught.
– Did
Nov 24 at 9:13
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@Did: I didn't want to cast any aspersions on the way complex analysis is taught. But it is a fact that, e.g., nobody really knows or cares what he is doing when using Wirtinger calculus.
– Christian Blatter
Nov 24 at 9:20
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
@ChristianBlatter conjugate$bar{z}$ is analytic but i said it is not in non trivial form...its in trivial form...
– neelkanth
Nov 25 at 7:53
add a comment |
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$(Re^2z+Im^2z)/overline z$? But seriously, we want $partial /partial overline z=0$
– Hagen von Eitzen
Nov 23 at 6:48
@HagenvonEitzen its in trivial form... its nothing but $z$ function.
– neelkanth
Nov 23 at 6:49
@HagenvonEitzen I am saying its in nontrivial form
– neelkanth
Nov 23 at 6:53