Mixing
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Are there multiple solutions to a fractional root/radical (over the complex numbers)?
$begingroup$
Any high school student would know that the square root of a real number other than 0 (let's say 7) has at most 2 real solutions: the principal root (in this case about 2.646) and -1 times whatever the principal root is (-2.646).
If we extend our solution range to the complex numbers, then cube roots, fourth roots, fifth roots and so on will also have multiple solutions.
For instance, if one tryed to to solve solutions for the cube root of 17 under the complex numbers, they would get 3 results:
- $^3√17 ≈ 2.571$
- $^3√17 ≈-1.286+2.227i$
- $^3√17 ≈-1.286-2.227i$
Likewise, if you were calculate all the answers to $^4√17$ you would end up with 4 solutions under the complex numbers:
$^4√17 ≈2.031$
$^4√17 ≈-2.031$
$^4√17 ≈-2.031i$
$^4√17 ≈2.031i$
These multiple-solution radicals are suited for $n$th roots where n is an integer and produce $n$ solutions.
The solutions themselves, here written in polar coordinates, go like this:
The 1st solution is of course the principal root $p$.
The 2nd solution is $p ∠(360/n)˚$
The 3rd solution is $p ∠(720/n)˚$
The 4th solution is $p ∠(1,080/n)˚$
The 5th solution is $p ∠(1,440/n)˚$
.
.
.
The jth solution is $p ∠(360(j-1)/n)˚$
where:
$n$ is the root degree used.
$j$ is the index number of the solution ($i$ is not used because it conflicts with the symbol for the imaginary unit)
$p$ is the principal root.
My question is that do $n$th roots with $n$ not being an integer (like $^pi√40$) also have multiple solutions?
(Note that, in this case, there will no longer be exactly $n$ solutions to the radical. This is because having a fractional number of solutions to a problem does not make sense in the real world.)
complex-numbers radicals
asked Jan 12 at 9:37
NirvanaNirvana
64
$endgroup$
add a comment |
$begingroup$
Any high school student would know that the square root of a real number other than 0 (let's say 7) has at most 2 real solutions: the principal root (in this case about 2.646) and -1 times whatever the principal root is (-2.646).
If we extend our solution range to the complex numbers, then cube roots, fourth roots, fifth roots and so on will also have multiple solutions.
For instance, if one tryed to to solve solutions for the cube root of 17 under the complex numbers, they would get 3 results:
- $^3√17 ≈ 2.571$
- $^3√17 ≈-1.286+2.227i$
- $^3√17 ≈-1.286-2.227i$
Likewise, if you were calculate all the answers to $^4√17$ you would end up with 4 solutions under the complex numbers:
$^4√17 ≈2.031$
$^4√17 ≈-2.031$
$^4√17 ≈-2.031i$
$^4√17 ≈2.031i$
These multiple-solution radicals are suited for $n$th roots where n is an integer and produce $n$ solutions.
The solutions themselves, here written in polar coordinates, go like this:
The 1st solution is of course the principal root $p$.
The 2nd solution is $p ∠(360/n)˚$
The 3rd solution is $p ∠(720/n)˚$
The 4th solution is $p ∠(1,080/n)˚$
The 5th solution is $p ∠(1,440/n)˚$
.
.
.
The jth solution is $p ∠(360(j-1)/n)˚$
where:
$n$ is the root degree used.
$j$ is the index number of the solution ($i$ is not used because it conflicts with the symbol for the imaginary unit)
$p$ is the principal root.
My question is that do $n$th roots with $n$ not being an integer (like $^pi√40$) also have multiple solutions?
(Note that, in this case, there will no longer be exactly $n$ solutions to the radical. This is because having a fractional number of solutions to a problem does not make sense in the real world.)
complex-numbers radicals
asked Jan 12 at 9:37
NirvanaNirvana
64
$endgroup$
add a comment |
$begingroup$
Any high school student would know that the square root of a real number other than 0 (let's say 7) has at most 2 real solutions: the principal root (in this case about 2.646) and -1 times whatever the principal root is (-2.646).
If we extend our solution range to the complex numbers, then cube roots, fourth roots, fifth roots and so on will also have multiple solutions.
For instance, if one tryed to to solve solutions for the cube root of 17 under the complex numbers, they would get 3 results:
- $^3√17 ≈ 2.571$
- $^3√17 ≈-1.286+2.227i$
- $^3√17 ≈-1.286-2.227i$
Likewise, if you were calculate all the answers to $^4√17$ you would end up with 4 solutions under the complex numbers:
$^4√17 ≈2.031$
$^4√17 ≈-2.031$
$^4√17 ≈-2.031i$
$^4√17 ≈2.031i$
These multiple-solution radicals are suited for $n$th roots where n is an integer and produce $n$ solutions.
The solutions themselves, here written in polar coordinates, go like this:
The 1st solution is of course the principal root $p$.
The 2nd solution is $p ∠(360/n)˚$
The 3rd solution is $p ∠(720/n)˚$
The 4th solution is $p ∠(1,080/n)˚$
The 5th solution is $p ∠(1,440/n)˚$
.
.
.
The jth solution is $p ∠(360(j-1)/n)˚$
where:
$n$ is the root degree used.
$j$ is the index number of the solution ($i$ is not used because it conflicts with the symbol for the imaginary unit)
$p$ is the principal root.
My question is that do $n$th roots with $n$ not being an integer (like $^pi√40$) also have multiple solutions?
(Note that, in this case, there will no longer be exactly $n$ solutions to the radical. This is because having a fractional number of solutions to a problem does not make sense in the real world.)
complex-numbers radicals
asked Jan 12 at 9:37
NirvanaNirvana
64
$endgroup$
Any high school student would know that the square root of a real number other than 0 (let's say 7) has at most 2 real solutions: the principal root (in this case about 2.646) and -1 times whatever the principal root is (-2.646).
If we extend our solution range to the complex numbers, then cube roots, fourth roots, fifth roots and so on will also have multiple solutions.
For instance, if one tryed to to solve solutions for the cube root of 17 under the complex numbers, they would get 3 results:
- $^3√17 ≈ 2.571$
- $^3√17 ≈-1.286+2.227i$
- $^3√17 ≈-1.286-2.227i$
Likewise, if you were calculate all the answers to $^4√17$ you would end up with 4 solutions under the complex numbers:
$^4√17 ≈2.031$
$^4√17 ≈-2.031$
$^4√17 ≈-2.031i$
$^4√17 ≈2.031i$
These multiple-solution radicals are suited for $n$th roots where n is an integer and produce $n$ solutions.
The solutions themselves, here written in polar coordinates, go like this:
The 1st solution is of course the principal root $p$.
The 2nd solution is $p ∠(360/n)˚$
The 3rd solution is $p ∠(720/n)˚$
The 4th solution is $p ∠(1,080/n)˚$
The 5th solution is $p ∠(1,440/n)˚$
.
.
.
The jth solution is $p ∠(360(j-1)/n)˚$
where:
$n$ is the root degree used.
$j$ is the index number of the solution ($i$ is not used because it conflicts with the symbol for the imaginary unit)
$p$ is the principal root.
My question is that do $n$th roots with $n$ not being an integer (like $^pi√40$) also have multiple solutions?
(Note that, in this case, there will no longer be exactly $n$ solutions to the radical. This is because having a fractional number of solutions to a problem does not make sense in the real world.)
complex-numbers radicals
complex-numbers radicals
asked Jan 12 at 9:37
NirvanaNirvana
64
asked Jan 12 at 9:37
NirvanaNirvana
64
edited Jan 12 at 20:38
Nirvana
asked Jan 12 at 9:37
NirvanaNirvana
64
asked Jan 12 at 9:37
NirvanaNirvana
64
asked Jan 12 at 9:37
NirvanaNirvana
64
64
add a comment |
add a comment |
1 Answer
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$begingroup$
In case the root being taken is a non-integer rational, then, the number of roots is still finite (and equal to the numerator of the reduced fraction form of the rational) because we can rewrite $sqrt[frac{a}{b}] cquad$ as $sqrt[a]{c^b}quad$- so, for instance, here you'd get $a$ roots. If the root being taken is irrational, then, you'll probably find infinitely many roots.
Suppose $e^{itheta}$ is an $n$th root ($n$ is irrational) of 1. Given that $e^{2pi i}=1$, any number of the form $e^{i(theta+kfrac{2pi}{n})}$, where $k$ is an integer, will also be an $n$th root of 1. Now suppose some number of the form $e^{i(theta+ufrac{2pi}{n})}$ equals some number of the form $e^{i(theta+vfrac{2pi}{n})}$, where $u$ and $v$ are distinct integers. Then, it must be the case that $(u-v)frac{2pi}{n}$ is some integer multiple of $2pi$ (referring to Euler's formula that $e^{ix}=cis(x)$). So $(u-v)frac{2pi}{n}=2pi mimplies n=(u-v)/m$. Since $u$ and $v$ were distinct, $m$ was non-zero- but that would imply that $n$ was rational, a contradiction. So there are infinitely many distinct irrational roots of unity (and by extension for all other elements of $mathbb C$ as well).
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
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1 Answer
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1 Answer
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$begingroup$
In case the root being taken is a non-integer rational, then, the number of roots is still finite (and equal to the numerator of the reduced fraction form of the rational) because we can rewrite $sqrt[frac{a}{b}] cquad$ as $sqrt[a]{c^b}quad$- so, for instance, here you'd get $a$ roots. If the root being taken is irrational, then, you'll probably find infinitely many roots.
Suppose $e^{itheta}$ is an $n$th root ($n$ is irrational) of 1. Given that $e^{2pi i}=1$, any number of the form $e^{i(theta+kfrac{2pi}{n})}$, where $k$ is an integer, will also be an $n$th root of 1. Now suppose some number of the form $e^{i(theta+ufrac{2pi}{n})}$ equals some number of the form $e^{i(theta+vfrac{2pi}{n})}$, where $u$ and $v$ are distinct integers. Then, it must be the case that $(u-v)frac{2pi}{n}$ is some integer multiple of $2pi$ (referring to Euler's formula that $e^{ix}=cis(x)$). So $(u-v)frac{2pi}{n}=2pi mimplies n=(u-v)/m$. Since $u$ and $v$ were distinct, $m$ was non-zero- but that would imply that $n$ was rational, a contradiction. So there are infinitely many distinct irrational roots of unity (and by extension for all other elements of $mathbb C$ as well).
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
$endgroup$
add a comment |
$begingroup$
In case the root being taken is a non-integer rational, then, the number of roots is still finite (and equal to the numerator of the reduced fraction form of the rational) because we can rewrite $sqrt[frac{a}{b}] cquad$ as $sqrt[a]{c^b}quad$- so, for instance, here you'd get $a$ roots. If the root being taken is irrational, then, you'll probably find infinitely many roots.
Suppose $e^{itheta}$ is an $n$th root ($n$ is irrational) of 1. Given that $e^{2pi i}=1$, any number of the form $e^{i(theta+kfrac{2pi}{n})}$, where $k$ is an integer, will also be an $n$th root of 1. Now suppose some number of the form $e^{i(theta+ufrac{2pi}{n})}$ equals some number of the form $e^{i(theta+vfrac{2pi}{n})}$, where $u$ and $v$ are distinct integers. Then, it must be the case that $(u-v)frac{2pi}{n}$ is some integer multiple of $2pi$ (referring to Euler's formula that $e^{ix}=cis(x)$). So $(u-v)frac{2pi}{n}=2pi mimplies n=(u-v)/m$. Since $u$ and $v$ were distinct, $m$ was non-zero- but that would imply that $n$ was rational, a contradiction. So there are infinitely many distinct irrational roots of unity (and by extension for all other elements of $mathbb C$ as well).
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
$endgroup$
add a comment |
$begingroup$
In case the root being taken is a non-integer rational, then, the number of roots is still finite (and equal to the numerator of the reduced fraction form of the rational) because we can rewrite $sqrt[frac{a}{b}] cquad$ as $sqrt[a]{c^b}quad$- so, for instance, here you'd get $a$ roots. If the root being taken is irrational, then, you'll probably find infinitely many roots.
Suppose $e^{itheta}$ is an $n$th root ($n$ is irrational) of 1. Given that $e^{2pi i}=1$, any number of the form $e^{i(theta+kfrac{2pi}{n})}$, where $k$ is an integer, will also be an $n$th root of 1. Now suppose some number of the form $e^{i(theta+ufrac{2pi}{n})}$ equals some number of the form $e^{i(theta+vfrac{2pi}{n})}$, where $u$ and $v$ are distinct integers. Then, it must be the case that $(u-v)frac{2pi}{n}$ is some integer multiple of $2pi$ (referring to Euler's formula that $e^{ix}=cis(x)$). So $(u-v)frac{2pi}{n}=2pi mimplies n=(u-v)/m$. Since $u$ and $v$ were distinct, $m$ was non-zero- but that would imply that $n$ was rational, a contradiction. So there are infinitely many distinct irrational roots of unity (and by extension for all other elements of $mathbb C$ as well).
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
$endgroup$
In case the root being taken is a non-integer rational, then, the number of roots is still finite (and equal to the numerator of the reduced fraction form of the rational) because we can rewrite $sqrt[frac{a}{b}] cquad$ as $sqrt[a]{c^b}quad$- so, for instance, here you'd get $a$ roots. If the root being taken is irrational, then, you'll probably find infinitely many roots.
Suppose $e^{itheta}$ is an $n$th root ($n$ is irrational) of 1. Given that $e^{2pi i}=1$, any number of the form $e^{i(theta+kfrac{2pi}{n})}$, where $k$ is an integer, will also be an $n$th root of 1. Now suppose some number of the form $e^{i(theta+ufrac{2pi}{n})}$ equals some number of the form $e^{i(theta+vfrac{2pi}{n})}$, where $u$ and $v$ are distinct integers. Then, it must be the case that $(u-v)frac{2pi}{n}$ is some integer multiple of $2pi$ (referring to Euler's formula that $e^{ix}=cis(x)$). So $(u-v)frac{2pi}{n}=2pi mimplies n=(u-v)/m$. Since $u$ and $v$ were distinct, $m$ was non-zero- but that would imply that $n$ was rational, a contradiction. So there are infinitely many distinct irrational roots of unity (and by extension for all other elements of $mathbb C$ as well).
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
answered Jan 12 at 23:30
Cardioid_Ass_22Cardioid_Ass_22
35314
35314
add a comment |
add a comment |
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