Order of a Complex Number?
Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.
What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?
I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.
Any help would be great, thanks.
abstract-algebra group-theory
add a comment |
Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.
What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?
I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.
Any help would be great, thanks.
abstract-algebra group-theory
I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
1
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago
add a comment |
Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.
What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?
I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.
Any help would be great, thanks.
abstract-algebra group-theory
Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.
What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?
I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.
Any help would be great, thanks.
abstract-algebra group-theory
abstract-algebra group-theory
edited 13 mins ago
Gaby Alfonso
676315
676315
asked 1 hour ago
Reety
12910
12910
I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
1
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago
add a comment |
I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
1
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago
I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
1
1
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago
add a comment |
4 Answers
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Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
add a comment |
If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.
$z$ has a finite order iff $|z| = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.
New contributor
add a comment |
Find the smallest natural $n$ so that $z^n = 1$.
$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$
So $z^n = e^{ifrac {2npi} 3}$
Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.
So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.
So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.
That's.... not a hard question.
add a comment |
Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$
Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$
(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$
(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$
(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$
Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t), $ so $z=|z|(cos t +isin t)=|z|e^{it}.$
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4 Answers
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Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
add a comment |
Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
add a comment |
Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?
Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?
edited 48 mins ago
answered 57 mins ago
Thomas Shelby
1,383216
1,383216
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
add a comment |
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
1
1
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Thank you very much, its when n=3 so if I'm correct this must mean it is of order 3?
– Reety
56 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
Yes. ${}{} {}{}{}$
– Thomas Shelby
50 mins ago
add a comment |
If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.
$z$ has a finite order iff $|z| = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.
New contributor
add a comment |
If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.
$z$ has a finite order iff $|z| = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.
New contributor
add a comment |
If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.
$z$ has a finite order iff $|z| = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.
New contributor
If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.
$z$ has a finite order iff $|z| = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.
New contributor
edited 1 min ago
Gaby Alfonso
676315
676315
New contributor
answered 51 mins ago
Zachary Hunter
111
111
New contributor
New contributor
add a comment |
add a comment |
Find the smallest natural $n$ so that $z^n = 1$.
$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$
So $z^n = e^{ifrac {2npi} 3}$
Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.
So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.
So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.
That's.... not a hard question.
add a comment |
Find the smallest natural $n$ so that $z^n = 1$.
$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$
So $z^n = e^{ifrac {2npi} 3}$
Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.
So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.
So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.
That's.... not a hard question.
add a comment |
Find the smallest natural $n$ so that $z^n = 1$.
$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$
So $z^n = e^{ifrac {2npi} 3}$
Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.
So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.
So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.
That's.... not a hard question.
Find the smallest natural $n$ so that $z^n = 1$.
$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$
So $z^n = e^{ifrac {2npi} 3}$
Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.
So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.
So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.
That's.... not a hard question.
answered 42 mins ago
fleablood
68.1k22684
68.1k22684
add a comment |
add a comment |
Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$
Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$
(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$
(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$
(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$
Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t), $ so $z=|z|(cos t +isin t)=|z|e^{it}.$
add a comment |
Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$
Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$
(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$
(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$
(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$
Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t), $ so $z=|z|(cos t +isin t)=|z|e^{it}.$
add a comment |
Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$
Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$
(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$
(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$
(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$
Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t), $ so $z=|z|(cos t +isin t)=|z|e^{it}.$
Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$
Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$
(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$
(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$
(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$
Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t), $ so $z=|z|(cos t +isin t)=|z|e^{it}.$
answered 1 min ago
DanielWainfleet
34k31647
34k31647
add a comment |
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I haven't heard of complex numbers being referred to as having any such property. Do you have any more context? What did previous parts of the question say for instance?
– John Doe
1 hour ago
This is the only previous part of the question, after it just states De Moivres Theorem. The aim of this exercise is to determine all elements of finite order in $C^{∗}$, the multiplicative group of non-zero complex numbers.@John Doe
– Reety
1 hour ago
1
Ahh ok - now that you mention group theory, this may make more sense. The order of a group element is the lowest (integer) power to which that element has to be raised for you to reach the identity. In your group, this is any $ainBbb C$ such that $a^n=1$ for $ninBbb N$. Then the order of your element $a$ will be $n$.
– John Doe
1 hour ago