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Find Where $f(z)=x^2+iy^2$ Differentiable And Analytic
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Let $f(z)=x^2+iy^2$ find where it is differentiable and where it is analytic in $mathbbC$
For a function to be differentiable at a point it should fulfil C-R equations, we have
$u(x,y)=x^2$ and $v(x,y)=y^2$
So $u_x=2x=2y=v_y$ and $u_y=0=0-v_x$
So the function is differentiable iff $2x=2yiff x=y$
So the function is differentiable on the straight line $y=x$
So $f(z)$ is not analytic? as it is only differentiable on a straight line and not in areas are a point? on the other hand the function seems to have no problematic points (singularities) so it is analytic?
complex-analysis analytic-functions
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
asked Jul 15 at 12:20
newhere
759310
add a comment |Â
up vote
0
down vote
favorite
Let $f(z)=x^2+iy^2$ find where it is differentiable and where it is analytic in $mathbbC$
For a function to be differentiable at a point it should fulfil C-R equations, we have
$u(x,y)=x^2$ and $v(x,y)=y^2$
So $u_x=2x=2y=v_y$ and $u_y=0=0-v_x$
So the function is differentiable iff $2x=2yiff x=y$
So the function is differentiable on the straight line $y=x$
So $f(z)$ is not analytic? as it is only differentiable on a straight line and not in areas are a point? on the other hand the function seems to have no problematic points (singularities) so it is analytic?
complex-analysis analytic-functions
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
asked Jul 15 at 12:20
newhere
759310
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f(z)=x^2+iy^2$ find where it is differentiable and where it is analytic in $mathbbC$
For a function to be differentiable at a point it should fulfil C-R equations, we have
$u(x,y)=x^2$ and $v(x,y)=y^2$
So $u_x=2x=2y=v_y$ and $u_y=0=0-v_x$
So the function is differentiable iff $2x=2yiff x=y$
So the function is differentiable on the straight line $y=x$
So $f(z)$ is not analytic? as it is only differentiable on a straight line and not in areas are a point? on the other hand the function seems to have no problematic points (singularities) so it is analytic?
complex-analysis analytic-functions
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
asked Jul 15 at 12:20
newhere
759310
Let $f(z)=x^2+iy^2$ find where it is differentiable and where it is analytic in $mathbbC$
For a function to be differentiable at a point it should fulfil C-R equations, we have
$u(x,y)=x^2$ and $v(x,y)=y^2$
So $u_x=2x=2y=v_y$ and $u_y=0=0-v_x$
So the function is differentiable iff $2x=2yiff x=y$
So the function is differentiable on the straight line $y=x$
So $f(z)$ is not analytic? as it is only differentiable on a straight line and not in areas are a point? on the other hand the function seems to have no problematic points (singularities) so it is analytic?
complex-analysis analytic-functions
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
asked Jul 15 at 12:20
newhere
759310
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
edited Jul 15 at 12:24
José Carlos Santos
114k1698177
114k1698177
asked Jul 15 at 12:20
newhere
759310
asked Jul 15 at 12:20
newhere
759310
asked Jul 15 at 12:20
newhere
759310
759310
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
 |Â
show 1 more comment
up vote
2
down vote
If you want differentiable as a function from $mathbbR^2$ to $mathbbR^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.
The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.
To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.
This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.
answered Jul 15 at 12:29
paracutey
212
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
That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
 |Â
show 1 more comment
up vote
2
down vote
accepted
That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
That depends on the definition of analytic function that you use. But if it is the usual one (which implies that if $f$ is analytic at a point $z$ then it is analytic in all points of a neighborhood of $z$) than $f$ is analytic nowhere.
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
answered Jul 15 at 12:24
José Carlos Santos
114k1698177
114k1698177
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
 |Â
show 1 more comment
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
Can we add that because $u_xx+u_vvneq 0$ and $v_xx+v_yyneq 0$ so the function is analytic nowhere?
– newhere
Jul 15 at 12:38
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
No. Why is that?
– José Carlos Santos
Jul 15 at 13:00
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
Because a function is analytic iff $u,v$ are harmonic conjugate
– newhere
Jul 15 at 13:16
1
1
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
Note that it was possible that $f$ was not analytic globally while, at the same time, it was analytic on some open subset of $mathbb C$.
– José Carlos Santos
Jul 15 at 13:18
1
1
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
Take$$f(z)=begincasesoperatornameRe^2z&text if operatornameRez>0\0&text otherwise.endcases$$Then the largest open subset of $mathbb C$ on which $f$ is analytic is $zinmathbbC,$.
– José Carlos Santos
Jul 15 at 13:59
 |Â
show 1 more comment
up vote
2
down vote
If you want differentiable as a function from $mathbbR^2$ to $mathbbR^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.
The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.
To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.
This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.
answered Jul 15 at 12:29
paracutey
212
add a comment |Â
up vote
2
down vote
If you want differentiable as a function from $mathbbR^2$ to $mathbbR^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.
The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.
To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.
This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.
answered Jul 15 at 12:29
paracutey
212
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you want differentiable as a function from $mathbbR^2$ to $mathbbR^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.
The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.
To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.
This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.
answered Jul 15 at 12:29
paracutey
212
If you want differentiable as a function from $mathbbR^2$ to $mathbbR^2$ ($f(x,y)=(x^2,y^2)$). This function is differentiable at all points, since its components have continuous partial derivatives everywhere.
The Cauchy-Riemann equations give you where it is complex-differentiable. You have found that complex derivative exist at every point of the line $y=x$.
To be analytic at a point it has to have complex derivative at a neighborhood of the point. This is because being analytic means that the function is equal to a convergent power series with a radius of convergence larger than $0$. In the interior of the disc of convergence, the sum of a power series has complex derivatives. Take into account that while this is a simple result, it is something that still requires a proof. As such, it is good to make it explicit in the proof since what you get from Cauchy-Riemann is existence of complex derivatives, a related, but different condition.
This function doesn't have complex derivative on any open set since the line $y=x$ has an empty interior. Therefore, it is not analytic anywhere.
answered Jul 15 at 12:29
paracutey
212
edited Jul 15 at 12:43
answered Jul 15 at 12:29
paracutey
212
answered Jul 15 at 12:29
paracutey
212
answered Jul 15 at 12:29
paracutey
212
212
add a comment |Â
add a comment |Â
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