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Why level curves of $z^2$ are not orthogonal for $u=v=0$?
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Theorem. Let the function $f(z) = u(x, y) + iv(x, y)$ be analytic in a domain $D$, and Let the families of level curves be $u(x, y) = c_1$ and $v(x, y) = c_2$, where $c_1$ and $c_2$ are arbitrary real constants. Then these families are orthogonal.
Regarding the mentioned Theorem the two following examples are confusing:
Consider $f(z)=z^2$. I plot the $x^2-y^2=a$ and $2xy=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 but it fails to be orthogonal! Does it contradict with the above theorem?
And, consider $f(z)=dfrac1z$. I plot the $x/(x^2+y^2)=a$ and $y/(x^2+y^2)=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 and still it doesn't fail to be orthogonal but $f(z)=dfrac1z$ is not analytic for $z=0$! Does it contradict with the above theorem?
complex-analysis examples-counterexamples
asked Jul 31 at 14:01
Edi
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Theorem. Let the function $f(z) = u(x, y) + iv(x, y)$ be analytic in a domain $D$, and Let the families of level curves be $u(x, y) = c_1$ and $v(x, y) = c_2$, where $c_1$ and $c_2$ are arbitrary real constants. Then these families are orthogonal.
Regarding the mentioned Theorem the two following examples are confusing:
Consider $f(z)=z^2$. I plot the $x^2-y^2=a$ and $2xy=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 but it fails to be orthogonal! Does it contradict with the above theorem?
And, consider $f(z)=dfrac1z$. I plot the $x/(x^2+y^2)=a$ and $y/(x^2+y^2)=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 and still it doesn't fail to be orthogonal but $f(z)=dfrac1z$ is not analytic for $z=0$! Does it contradict with the above theorem?
complex-analysis examples-counterexamples
asked Jul 31 at 14:01
Edi
1,121728
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Theorem. Let the function $f(z) = u(x, y) + iv(x, y)$ be analytic in a domain $D$, and Let the families of level curves be $u(x, y) = c_1$ and $v(x, y) = c_2$, where $c_1$ and $c_2$ are arbitrary real constants. Then these families are orthogonal.
Regarding the mentioned Theorem the two following examples are confusing:
Consider $f(z)=z^2$. I plot the $x^2-y^2=a$ and $2xy=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 but it fails to be orthogonal! Does it contradict with the above theorem?
And, consider $f(z)=dfrac1z$. I plot the $x/(x^2+y^2)=a$ and $y/(x^2+y^2)=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 and still it doesn't fail to be orthogonal but $f(z)=dfrac1z$ is not analytic for $z=0$! Does it contradict with the above theorem?
complex-analysis examples-counterexamples
asked Jul 31 at 14:01
Edi
1,121728
Theorem. Let the function $f(z) = u(x, y) + iv(x, y)$ be analytic in a domain $D$, and Let the families of level curves be $u(x, y) = c_1$ and $v(x, y) = c_2$, where $c_1$ and $c_2$ are arbitrary real constants. Then these families are orthogonal.
Regarding the mentioned Theorem the two following examples are confusing:
Consider $f(z)=z^2$. I plot the $x^2-y^2=a$ and $2xy=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 but it fails to be orthogonal! Does it contradict with the above theorem?
And, consider $f(z)=dfrac1z$. I plot the $x/(x^2+y^2)=a$ and $y/(x^2+y^2)=b$ for different values of a and b and they are orthogonal and I tried for a=b=0 and still it doesn't fail to be orthogonal but $f(z)=dfrac1z$ is not analytic for $z=0$! Does it contradict with the above theorem?
complex-analysis examples-counterexamples
asked Jul 31 at 14:01
Edi
1,121728
asked Jul 31 at 14:01
Edi
1,121728
asked Jul 31 at 14:01
Edi
1,121728
asked Jul 31 at 14:01
Edi
1,121728
1,121728
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
4
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The theorem is only valied for those points at which $f'$ is not $0$. The explains your first question.
Concerning the second one, the function isn't even defined at $0$.
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
 |Â
show 1 more comment
up vote
0
down vote
The curves $x^2-y^2=a$ and $xy=b$ are two pencils of equilateral hyperbolas (actually the same pencils, rotated by $45°$).
You can observe that all these curves are orthogonal to each other, with a single exception: $a=b=0$. But the world is safe because at the intersection point $(z=0$), the curves don't have a unique direction and the property has no meaning there.
answered Jul 31 at 14:28
Yves Daoust
110k665203
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The theorem is only valied for those points at which $f'$ is not $0$. The explains your first question.
Concerning the second one, the function isn't even defined at $0$.
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
 |Â
show 1 more comment
up vote
4
down vote
accepted
The theorem is only valied for those points at which $f'$ is not $0$. The explains your first question.
Concerning the second one, the function isn't even defined at $0$.
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The theorem is only valied for those points at which $f'$ is not $0$. The explains your first question.
Concerning the second one, the function isn't even defined at $0$.
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
The theorem is only valied for those points at which $f'$ is not $0$. The explains your first question.
Concerning the second one, the function isn't even defined at $0$.
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
edited Jul 31 at 14:15
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
answered Jul 31 at 14:03
José Carlos Santos
112k1696172
112k1696172
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
 |Â
show 1 more comment
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
Book doesn't proved a proof so I proved the theorem myself using Dini's Theorem for $u$ and $v$ (in real analysis as they are real) which doesn't require for $u$ and $v$ to be non-zero. How so "The theorem is only valid for those points at which f is not 0"?
– Edi
Jul 31 at 14:13
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
@Edi My mistake: I meant $f'$ is not zero. I've edited my answer. See here, for instance.
– José Carlos Santos
Jul 31 at 14:16
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
So if $f'$ is zero then $u_x=v_x=0$ and thus $u_y=v_y=0$ so a condition Dini's Theorem fails to satisfy, am I right?
– Edi
Jul 31 at 14:19
1
1
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
@Edi In order for me to answer that question, you'll have to tell me which theoreom is that. It can't be the one I know.
– José Carlos Santos
Jul 31 at 15:24
1
1
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
@Edi I'm glad I could help.
– José Carlos Santos
Jul 31 at 15:49
 |Â
show 1 more comment
up vote
0
down vote
The curves $x^2-y^2=a$ and $xy=b$ are two pencils of equilateral hyperbolas (actually the same pencils, rotated by $45°$).
You can observe that all these curves are orthogonal to each other, with a single exception: $a=b=0$. But the world is safe because at the intersection point $(z=0$), the curves don't have a unique direction and the property has no meaning there.
answered Jul 31 at 14:28
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
The curves $x^2-y^2=a$ and $xy=b$ are two pencils of equilateral hyperbolas (actually the same pencils, rotated by $45°$).
You can observe that all these curves are orthogonal to each other, with a single exception: $a=b=0$. But the world is safe because at the intersection point $(z=0$), the curves don't have a unique direction and the property has no meaning there.
answered Jul 31 at 14:28
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The curves $x^2-y^2=a$ and $xy=b$ are two pencils of equilateral hyperbolas (actually the same pencils, rotated by $45°$).
You can observe that all these curves are orthogonal to each other, with a single exception: $a=b=0$. But the world is safe because at the intersection point $(z=0$), the curves don't have a unique direction and the property has no meaning there.
answered Jul 31 at 14:28
Yves Daoust
110k665203
The curves $x^2-y^2=a$ and $xy=b$ are two pencils of equilateral hyperbolas (actually the same pencils, rotated by $45°$).
You can observe that all these curves are orthogonal to each other, with a single exception: $a=b=0$. But the world is safe because at the intersection point $(z=0$), the curves don't have a unique direction and the property has no meaning there.
answered Jul 31 at 14:28
Yves Daoust
110k665203
answered Jul 31 at 14:28
Yves Daoust
110k665203
answered Jul 31 at 14:28
Yves Daoust
110k665203
answered Jul 31 at 14:28
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
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