Mixing
How do you determine the principal root of a unit complex number?
$begingroup$
Let's suppose t be in the inverval $(-pi, pi]$ and that $n$ is a natural number. What is $(cos t + isin t )^{frac 1n}$? Using Euler's formula would give us the following:
$(cos t + isin t )^{frac 1n}=$
$(e^{it})^{frac 1n}=$
$e^{ittimes frac 1n} = $
$e^{frac{it}{n}} = $
$cos frac 1n + isin frac 1n$.
However, this would be problematic if we're taking on odd root of $-1$. When we're just dealing with the real numbers, any odd root of $-1$ is $-1$. However, if $n$ is odd, then based on formula above, since the argument of $-1$ is $pi$, $(-1)^{frac 1n} = cos frac{pi}{n}+ isin frac{pi}{n}$. So $(-1)^{frac 13}$ would be equal to $ frac 12 + i frac{sqrt 3}{2}$, even though it would just be $-1$ if we were only dealing with the real numbers. It doesn't make sense that the same operation would yield a different result just because we've extended the number-system we're working with. So, I was wondering if this in fact, is the correct formula for determining the principal root of a unti complex number.
$(cos t + isin t )^{frac 1n}= cos frac 1n + isin frac 1n$ if $-pi<t<pi$ or $n$ is even
$=-1$ if $t=pi$ and $n$ is odd
If this isn't the correct formula, then what is the correct formula?
complex-numbers
edited Jan 31 at 15:34
Martigan
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
$endgroup$
add a comment |
$begingroup$
Let's suppose t be in the inverval $(-pi, pi]$ and that $n$ is a natural number. What is $(cos t + isin t )^{frac 1n}$? Using Euler's formula would give us the following:
$(cos t + isin t )^{frac 1n}=$
$(e^{it})^{frac 1n}=$
$e^{ittimes frac 1n} = $
$e^{frac{it}{n}} = $
$cos frac 1n + isin frac 1n$.
However, this would be problematic if we're taking on odd root of $-1$. When we're just dealing with the real numbers, any odd root of $-1$ is $-1$. However, if $n$ is odd, then based on formula above, since the argument of $-1$ is $pi$, $(-1)^{frac 1n} = cos frac{pi}{n}+ isin frac{pi}{n}$. So $(-1)^{frac 13}$ would be equal to $ frac 12 + i frac{sqrt 3}{2}$, even though it would just be $-1$ if we were only dealing with the real numbers. It doesn't make sense that the same operation would yield a different result just because we've extended the number-system we're working with. So, I was wondering if this in fact, is the correct formula for determining the principal root of a unti complex number.
$(cos t + isin t )^{frac 1n}= cos frac 1n + isin frac 1n$ if $-pi<t<pi$ or $n$ is even
$=-1$ if $t=pi$ and $n$ is odd
If this isn't the correct formula, then what is the correct formula?
complex-numbers
edited Jan 31 at 15:34
Martigan
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
$endgroup$
$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28
add a comment |
$begingroup$
Let's suppose t be in the inverval $(-pi, pi]$ and that $n$ is a natural number. What is $(cos t + isin t )^{frac 1n}$? Using Euler's formula would give us the following:
$(cos t + isin t )^{frac 1n}=$
$(e^{it})^{frac 1n}=$
$e^{ittimes frac 1n} = $
$e^{frac{it}{n}} = $
$cos frac 1n + isin frac 1n$.
However, this would be problematic if we're taking on odd root of $-1$. When we're just dealing with the real numbers, any odd root of $-1$ is $-1$. However, if $n$ is odd, then based on formula above, since the argument of $-1$ is $pi$, $(-1)^{frac 1n} = cos frac{pi}{n}+ isin frac{pi}{n}$. So $(-1)^{frac 13}$ would be equal to $ frac 12 + i frac{sqrt 3}{2}$, even though it would just be $-1$ if we were only dealing with the real numbers. It doesn't make sense that the same operation would yield a different result just because we've extended the number-system we're working with. So, I was wondering if this in fact, is the correct formula for determining the principal root of a unti complex number.
$(cos t + isin t )^{frac 1n}= cos frac 1n + isin frac 1n$ if $-pi<t<pi$ or $n$ is even
$=-1$ if $t=pi$ and $n$ is odd
If this isn't the correct formula, then what is the correct formula?
complex-numbers
edited Jan 31 at 15:34
Martigan
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
$endgroup$
Let's suppose t be in the inverval $(-pi, pi]$ and that $n$ is a natural number. What is $(cos t + isin t )^{frac 1n}$? Using Euler's formula would give us the following:
$(cos t + isin t )^{frac 1n}=$
$(e^{it})^{frac 1n}=$
$e^{ittimes frac 1n} = $
$e^{frac{it}{n}} = $
$cos frac 1n + isin frac 1n$.
However, this would be problematic if we're taking on odd root of $-1$. When we're just dealing with the real numbers, any odd root of $-1$ is $-1$. However, if $n$ is odd, then based on formula above, since the argument of $-1$ is $pi$, $(-1)^{frac 1n} = cos frac{pi}{n}+ isin frac{pi}{n}$. So $(-1)^{frac 13}$ would be equal to $ frac 12 + i frac{sqrt 3}{2}$, even though it would just be $-1$ if we were only dealing with the real numbers. It doesn't make sense that the same operation would yield a different result just because we've extended the number-system we're working with. So, I was wondering if this in fact, is the correct formula for determining the principal root of a unti complex number.
$(cos t + isin t )^{frac 1n}= cos frac 1n + isin frac 1n$ if $-pi<t<pi$ or $n$ is even
$=-1$ if $t=pi$ and $n$ is odd
If this isn't the correct formula, then what is the correct formula?
complex-numbers
complex-numbers
edited Jan 31 at 15:34
Martigan
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
edited Jan 31 at 15:34
Martigan
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
edited Jan 31 at 15:34
Martigan
5,223917
edited Jan 31 at 15:34
Martigan
5,223917
edited Jan 31 at 15:34
Martigan
5,223917
5,223917
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
asked Jan 31 at 15:23
Wire BowlWire Bowl
113
113
$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28
add a comment |
$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28
$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28
$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $frac{1}{2}+ifrac{sqrt{3}}{2},-1$ and $frac{1}{2}-ifrac{sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $frac{1}{2}+ifrac{sqrt{3}}{2}$.
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.
The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $mathbb{R}.$ since there aren't really notions of positive and negative in the same way.
One could do something by forcing all the angles of the roots between $[0,2pi)$ and then select the root with the smallest angle
answered Jan 31 at 15:49
marlowmarlow
254
$endgroup$
add a comment |
$begingroup$
For the n-th root there are n complex solutions, to visualize it see the picture.
answered Jan 31 at 17:37
jjzunjjzun
235
$endgroup$
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $frac{1}{2}+ifrac{sqrt{3}}{2},-1$ and $frac{1}{2}-ifrac{sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $frac{1}{2}+ifrac{sqrt{3}}{2}$.
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $frac{1}{2}+ifrac{sqrt{3}}{2},-1$ and $frac{1}{2}-ifrac{sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $frac{1}{2}+ifrac{sqrt{3}}{2}$.
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
$endgroup$
add a comment |
$begingroup$
This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $frac{1}{2}+ifrac{sqrt{3}}{2},-1$ and $frac{1}{2}-ifrac{sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $frac{1}{2}+ifrac{sqrt{3}}{2}$.
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
$endgroup$
This is the reason why one should never talk about "the" root of a number when dealing with complex numbers -1 has 3 cube roots, $frac{1}{2}+ifrac{sqrt{3}}{2},-1$ and $frac{1}{2}-ifrac{sqrt{3}}{2}$. Your method only gives the root with the smallest argument which is $frac{1}{2}+ifrac{sqrt{3}}{2}$.
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 15:42
Floris ClaassensFloris Claassens
1,33229
1,33229
add a comment |
add a comment |
$begingroup$
I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.
The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $mathbb{R}.$ since there aren't really notions of positive and negative in the same way.
One could do something by forcing all the angles of the roots between $[0,2pi)$ and then select the root with the smallest angle
answered Jan 31 at 15:49
marlowmarlow
254
$endgroup$
add a comment |
$begingroup$
I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.
The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $mathbb{R}.$ since there aren't really notions of positive and negative in the same way.
One could do something by forcing all the angles of the roots between $[0,2pi)$ and then select the root with the smallest angle
answered Jan 31 at 15:49
marlowmarlow
254
$endgroup$
add a comment |
$begingroup$
I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.
The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $mathbb{R}.$ since there aren't really notions of positive and negative in the same way.
One could do something by forcing all the angles of the roots between $[0,2pi)$ and then select the root with the smallest angle
answered Jan 31 at 15:49
marlowmarlow
254
$endgroup$
I could be wrong, but I don't think there is a universally accepted definition of "principal root" in the context of complex analysis. Since we are working over an algebraically closed field, $mathbb{C},$ the equation $z^n-z_0=0$ for $z_0$ having unit radius (or any modulus for that matter) has $n$ solutions.
The selection of one of these $n$-th roots as being special somehow is more arbitrary that (e.g.) the selection of a square root of a positive number in $mathbb{R}.$ since there aren't really notions of positive and negative in the same way.
One could do something by forcing all the angles of the roots between $[0,2pi)$ and then select the root with the smallest angle
answered Jan 31 at 15:49
marlowmarlow
254
answered Jan 31 at 15:49
marlowmarlow
254
answered Jan 31 at 15:49
marlowmarlow
254
answered Jan 31 at 15:49
marlowmarlow
254
254
add a comment |
add a comment |
$begingroup$
For the n-th root there are n complex solutions, to visualize it see the picture.
answered Jan 31 at 17:37
jjzunjjzun
235
$endgroup$
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
add a comment |
$begingroup$
For the n-th root there are n complex solutions, to visualize it see the picture.
answered Jan 31 at 17:37
jjzunjjzun
235
$endgroup$
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
add a comment |
$begingroup$
For the n-th root there are n complex solutions, to visualize it see the picture.
answered Jan 31 at 17:37
jjzunjjzun
235
$endgroup$
For the n-th root there are n complex solutions, to visualize it see the picture.
answered Jan 31 at 17:37
jjzunjjzun
235
answered Jan 31 at 17:37
jjzunjjzun
235
answered Jan 31 at 17:37
jjzunjjzun
235
answered Jan 31 at 17:37
jjzunjjzun
235
235
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
add a comment |
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
$begingroup$
I understand that. For instance, z^2 = 1 has 2 solutions: x = -1 and x = 1. Similarly, z^n = cos(t) + sin(t)i has n solutions; what I'm wondering is what (cos(t) + sin(t)i)^(1/n) is equal to.
$endgroup$
– Wire Bowl
Jan 31 at 19:40
add a comment |
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$begingroup$
Please use MathJax
$endgroup$
– J. W. Tanner
Jan 31 at 15:28