Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Artin chapter 2 exercisesLet $f:G longrightarrow G'$ be a group homomorphism. Prove that $operatornameIm(f)$ is a subgroup of $G'$.Prove that the composition of two group homomorphisms is a group homomorphism.Prove that the center of a group is normal.Prove that if $f:G longrightarrow G'$ is a group homomorphism, $f(x)^-1 = f(x^-1)$ for all $x in G$Let $f : mathbb Zto mathbb Z/xmathbb Z times mathbb Z/ymathbb Z$ be the homomorphism defined by $f (n) = (n + xZ, n + yZ)$…Let $X$ be a metric space with metric $d$. Show that $d:X times X longrightarrow mathbbR$ is continuous.Let $theta colon mathbbZlongrightarrowmathbbZtimesmathbbZ$ be a nonzero ring homomorphism. Prove that $kertheta = 0$.Show that the $mathbbQ$-homomorphism is well-defined.Homomorphism $f: mathbbZ_12 longrightarrow mathbbZ_30$
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Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Artin chapter 2 exercisesLet $f:G longrightarrow G'$ be a group homomorphism. Prove that $operatornameIm(f)$ is a subgroup of $G'$.Prove that the composition of two group homomorphisms is a group homomorphism.Prove that the center of a group is normal.Prove that if $f:G longrightarrow G'$ is a group homomorphism, $f(x)^-1 = f(x^-1)$ for all $x in G$Let $f : mathbb Zto mathbb Z/xmathbb Z times mathbb Z/ymathbb Z$ be the homomorphism defined by $f (n) = (n + xZ, n + yZ)$…Let $X$ be a metric space with metric $d$. Show that $d:X times X longrightarrow mathbbR$ is continuous.Let $theta colon mathbbZlongrightarrowmathbbZtimesmathbbZ$ be a nonzero ring homomorphism. Prove that $kertheta = 0$.Show that the $mathbbQ$-homomorphism is well-defined.Homomorphism $f: mathbbZ_12 longrightarrow mathbbZ_30$
$begingroup$
Can someone please verify my proof?
Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism, and determine its kernel and image.
Let $x$ and $y$ be arbitrary elements of $mathbbR^+$. Then,
begineqnarray
f(x+y) &=& e^i(x+y) \
&=& e^ixe^iy \
&=& f(x)times f(y)
endeqnarray
Also,
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR^+ \
&=& =1
endeqnarray
And,
begineqnarray
operatornameker(f)&=&x in mathbbR^+:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
abstract-algebra proof-verification
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
$endgroup$
add a comment |
$begingroup$
Can someone please verify my proof?
Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism, and determine its kernel and image.
Let $x$ and $y$ be arbitrary elements of $mathbbR^+$. Then,
begineqnarray
f(x+y) &=& e^i(x+y) \
&=& e^ixe^iy \
&=& f(x)times f(y)
endeqnarray
Also,
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR^+ \
&=& =1
endeqnarray
And,
begineqnarray
operatornameker(f)&=&x in mathbbR^+:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
abstract-algebra proof-verification
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
$endgroup$
1
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39
add a comment |
$begingroup$
Can someone please verify my proof?
Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism, and determine its kernel and image.
Let $x$ and $y$ be arbitrary elements of $mathbbR^+$. Then,
begineqnarray
f(x+y) &=& e^i(x+y) \
&=& e^ixe^iy \
&=& f(x)times f(y)
endeqnarray
Also,
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR^+ \
&=& =1
endeqnarray
And,
begineqnarray
operatornameker(f)&=&x in mathbbR^+:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
abstract-algebra proof-verification
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
$endgroup$
Can someone please verify my proof?
Show that $f:mathbbR^+ longrightarrow mathbbC^times$ defined by $f(x)=e^ix$ is a homomorphism, and determine its kernel and image.
Let $x$ and $y$ be arbitrary elements of $mathbbR^+$. Then,
begineqnarray
f(x+y) &=& e^i(x+y) \
&=& e^ixe^iy \
&=& f(x)times f(y)
endeqnarray
Also,
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR^+ \
&=& =1
endeqnarray
And,
begineqnarray
operatornameker(f)&=&x in mathbbR^+:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
abstract-algebra proof-verification
abstract-algebra proof-verification
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
edited Jun 3 '14 at 9:48
Shaun
10.5k113687
10.5k113687
asked Jun 3 '14 at 9:19
user154185user154185
723724
asked Jun 3 '14 at 9:19
user154185user154185
723724
asked Jun 3 '14 at 9:19
user154185user154185
723724
723724
1
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39
add a comment |
1
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39
1
1
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Perhaps you mean $(Bbb R,+)$ and $(Bbb C^times,cdot)$; but as written now there is a problem which bothers me, namely $Bbb R^+$ often denotes the positive elements of $Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2npi$ where $ninBbb Z$? and why is the image of $f$ exactly $xinBbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:mathbbR^+ longrightarrow mathbbC^times$
means
$f:(mathbbR,+) longrightarrow (mathbbC,times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR \
&=& x in mathbbC
endeqnarray
and
begineqnarray
operatornameker(f)&=&x in mathbbR:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
answered Aug 5 '18 at 18:13
hcharhchar
887
$endgroup$
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Perhaps you mean $(Bbb R,+)$ and $(Bbb C^times,cdot)$; but as written now there is a problem which bothers me, namely $Bbb R^+$ often denotes the positive elements of $Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2npi$ where $ninBbb Z$? and why is the image of $f$ exactly $xinBbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
Perhaps you mean $(Bbb R,+)$ and $(Bbb C^times,cdot)$; but as written now there is a problem which bothers me, namely $Bbb R^+$ often denotes the positive elements of $Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2npi$ where $ninBbb Z$? and why is the image of $f$ exactly $xinBbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
Perhaps you mean $(Bbb R,+)$ and $(Bbb C^times,cdot)$; but as written now there is a problem which bothers me, namely $Bbb R^+$ often denotes the positive elements of $Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2npi$ where $ninBbb Z$? and why is the image of $f$ exactly $xinBbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
Perhaps you mean $(Bbb R,+)$ and $(Bbb C^times,cdot)$; but as written now there is a problem which bothers me, namely $Bbb R^+$ often denotes the positive elements of $Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2npi$ where $ninBbb Z$? and why is the image of $f$ exactly $xinBbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
answered Jun 3 '14 at 10:00
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
add a comment |
add a comment |
$begingroup$
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:mathbbR^+ longrightarrow mathbbC^times$
means
$f:(mathbbR,+) longrightarrow (mathbbC,times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR \
&=& x in mathbbC
endeqnarray
and
begineqnarray
operatornameker(f)&=&x in mathbbR:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
answered Aug 5 '18 at 18:13
hcharhchar
887
$endgroup$
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
add a comment |
$begingroup$
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:mathbbR^+ longrightarrow mathbbC^times$
means
$f:(mathbbR,+) longrightarrow (mathbbC,times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR \
&=& x in mathbbC
endeqnarray
and
begineqnarray
operatornameker(f)&=&x in mathbbR:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
answered Aug 5 '18 at 18:13
hcharhchar
887
$endgroup$
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
add a comment |
$begingroup$
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:mathbbR^+ longrightarrow mathbbC^times$
means
$f:(mathbbR,+) longrightarrow (mathbbC,times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR \
&=& x in mathbbC
endeqnarray
and
begineqnarray
operatornameker(f)&=&x in mathbbR:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
answered Aug 5 '18 at 18:13
hcharhchar
887
$endgroup$
This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin. I believe the notation there
$f:mathbbR^+ longrightarrow mathbbC^times$
means
$f:(mathbbR,+) longrightarrow (mathbbC,times)$
In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
begineqnarray
operatornameIm(f) &=& e^ix:x in mathbbR \
&=& x in mathbbC
endeqnarray
and
begineqnarray
operatornameker(f)&=&x in mathbbR:e^ix=1 \
&=& 2 pi n: n in mathbbZ
endeqnarray
answered Aug 5 '18 at 18:13
hcharhchar
887
answered Aug 5 '18 at 18:13
hcharhchar
887
answered Aug 5 '18 at 18:13
hcharhchar
887
answered Aug 5 '18 at 18:13
hcharhchar
887
887
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
add a comment |
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
$begingroup$
That is not the image. The image is the circle of radius 1 in the complex plane.
$endgroup$
– Gunnar Sveinsson
Mar 31 at 6:42
add a comment |
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1
$begingroup$
Is there a specific point you weren't sure about? Your proof looks good to me.
$endgroup$
– Najib Idrissi
Jun 3 '14 at 9:38
$begingroup$
Your proof is correct. I've thought it was also necessary to check that $f(0) = 1+0i$ but I was wrong.
$endgroup$
– Bman72
Jun 3 '14 at 9:44
$begingroup$
Could you please clarify both Groups. I've read it as the set of Real Numbers under it's definition of addition with the set of Complex Numbers under it's definition of multiplication. Is that correct?
$endgroup$
– user150203
Mar 29 '18 at 5:39