A function $ f : [0,1] to BbbC $ is differentiable with derivative 0, is constant.
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In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and $ f : G to BbbC $ such that $f'(z)=0 forall z in G$ then f is constant"
He fixes some $z_0 in G$ and defines the set $A=$ $z in G: f(z)=f(z_0)$ and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) , t in [0,1]$ for some fixed $a in A $ and $epsilon >0$ such that $zin B(a,epsilon)subset G$ He wants to show that $B(a,epsilon) subset A.$
By proving
$g'(t)=0 $on $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.
proof-explanation
asked Aug 22 at 7:14
Uday Khanna
1577
add a comment |Â
up vote
0
down vote
favorite
In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and $ f : G to BbbC $ such that $f'(z)=0 forall z in G$ then f is constant"
He fixes some $z_0 in G$ and defines the set $A=$ $z in G: f(z)=f(z_0)$ and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) , t in [0,1]$ for some fixed $a in A $ and $epsilon >0$ such that $zin B(a,epsilon)subset G$ He wants to show that $B(a,epsilon) subset A.$
By proving
$g'(t)=0 $on $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.
proof-explanation
asked Aug 22 at 7:14
Uday Khanna
1577
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and $ f : G to BbbC $ such that $f'(z)=0 forall z in G$ then f is constant"
He fixes some $z_0 in G$ and defines the set $A=$ $z in G: f(z)=f(z_0)$ and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) , t in [0,1]$ for some fixed $a in A $ and $epsilon >0$ such that $zin B(a,epsilon)subset G$ He wants to show that $B(a,epsilon) subset A.$
By proving
$g'(t)=0 $on $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.
proof-explanation
asked Aug 22 at 7:14
Uday Khanna
1577
In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and $ f : G to BbbC $ such that $f'(z)=0 forall z in G$ then f is constant"
He fixes some $z_0 in G$ and defines the set $A=$ $z in G: f(z)=f(z_0)$ and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) , t in [0,1]$ for some fixed $a in A $ and $epsilon >0$ such that $zin B(a,epsilon)subset G$ He wants to show that $B(a,epsilon) subset A.$
By proving
$g'(t)=0 $on $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.
proof-explanation
asked Aug 22 at 7:14
Uday Khanna
1577
edited Aug 22 at 7:18
asked Aug 22 at 7:14
Uday Khanna
1577
asked Aug 22 at 7:14
Uday Khanna
1577
asked Aug 22 at 7:14
Uday Khanna
1577
1577
add a comment |Â
add a comment |Â
1 Answer
1
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0
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accepted
If $g:[0,1]to mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If $g:[0,1]to mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
add a comment |Â
up vote
0
down vote
accepted
If $g:[0,1]to mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If $g:[0,1]to mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
If $g:[0,1]to mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
answered Aug 22 at 7:17
Kavi Rama Murthy
23.4k2933
23.4k2933
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
add a comment |Â
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
Okay, to clarify, we observe $g(t): [0,1] to BbbR^2$ where if $g(t)=u(t)+i v(t)$ then $g'(t)=u'(t)+iv'(t) implies u $ and $ v$ are constant.
– Uday Khanna
Aug 22 at 7:48
add a comment |Â
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