Shouldn't the product of all $n$-th roots of unity be $1$?
$begingroup$
nth root of unity:
$$
1^{1/n} = e^{i2pi k/n} k=0,1,...,n-1
$$
multiplying together all the nth roots of unit to get back "1":
$$
1=prod_{k=0}^{n-1}e^{ileft(2pi k/nright)}
$$
for n=2:
$$
1=left(e^{i2pileft(0right)/2}right)left(e^{i2pileft(1right)/2}right) \
1=left(1right)left(-1right) \
1= -1
$$
why is 1 = -1?
For n=4:
$$
1=left(e^{i2pileft(0right)/4}right)left(e^{i2pileft(1right)/4}right)left(e^{i2pileft(2right)/4}right)left(e^{i2pileft(3right)/4}right) \
1=left(1right)left(iright)left(-1right)left(-iright) \
1= -1
$$
why is 1 = -1? shouldn't it be "1"?
A few definitions:
$$
e^{itheta} = cos(theta) + i sin(theta)
$$
$$
e^{-itheta} = cos(theta) - i sin(theta)
$$
$$
i*i = -1
$$
$$
frac{1}{i} = -i
$$
complex-numbers
edited Jan 13 at 2:33
Blue
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
$endgroup$
|
show 4 more comments
$begingroup$
nth root of unity:
$$
1^{1/n} = e^{i2pi k/n} k=0,1,...,n-1
$$
multiplying together all the nth roots of unit to get back "1":
$$
1=prod_{k=0}^{n-1}e^{ileft(2pi k/nright)}
$$
for n=2:
$$
1=left(e^{i2pileft(0right)/2}right)left(e^{i2pileft(1right)/2}right) \
1=left(1right)left(-1right) \
1= -1
$$
why is 1 = -1?
For n=4:
$$
1=left(e^{i2pileft(0right)/4}right)left(e^{i2pileft(1right)/4}right)left(e^{i2pileft(2right)/4}right)left(e^{i2pileft(3right)/4}right) \
1=left(1right)left(iright)left(-1right)left(-iright) \
1= -1
$$
why is 1 = -1? shouldn't it be "1"?
A few definitions:
$$
e^{itheta} = cos(theta) + i sin(theta)
$$
$$
e^{-itheta} = cos(theta) - i sin(theta)
$$
$$
i*i = -1
$$
$$
frac{1}{i} = -i
$$
complex-numbers
edited Jan 13 at 2:33
Blue
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
$endgroup$
7
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
3
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
3
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
2
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
2
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31
|
show 4 more comments
$begingroup$
nth root of unity:
$$
1^{1/n} = e^{i2pi k/n} k=0,1,...,n-1
$$
multiplying together all the nth roots of unit to get back "1":
$$
1=prod_{k=0}^{n-1}e^{ileft(2pi k/nright)}
$$
for n=2:
$$
1=left(e^{i2pileft(0right)/2}right)left(e^{i2pileft(1right)/2}right) \
1=left(1right)left(-1right) \
1= -1
$$
why is 1 = -1?
For n=4:
$$
1=left(e^{i2pileft(0right)/4}right)left(e^{i2pileft(1right)/4}right)left(e^{i2pileft(2right)/4}right)left(e^{i2pileft(3right)/4}right) \
1=left(1right)left(iright)left(-1right)left(-iright) \
1= -1
$$
why is 1 = -1? shouldn't it be "1"?
A few definitions:
$$
e^{itheta} = cos(theta) + i sin(theta)
$$
$$
e^{-itheta} = cos(theta) - i sin(theta)
$$
$$
i*i = -1
$$
$$
frac{1}{i} = -i
$$
complex-numbers
edited Jan 13 at 2:33
Blue
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
$endgroup$
nth root of unity:
$$
1^{1/n} = e^{i2pi k/n} k=0,1,...,n-1
$$
multiplying together all the nth roots of unit to get back "1":
$$
1=prod_{k=0}^{n-1}e^{ileft(2pi k/nright)}
$$
for n=2:
$$
1=left(e^{i2pileft(0right)/2}right)left(e^{i2pileft(1right)/2}right) \
1=left(1right)left(-1right) \
1= -1
$$
why is 1 = -1?
For n=4:
$$
1=left(e^{i2pileft(0right)/4}right)left(e^{i2pileft(1right)/4}right)left(e^{i2pileft(2right)/4}right)left(e^{i2pileft(3right)/4}right) \
1=left(1right)left(iright)left(-1right)left(-iright) \
1= -1
$$
why is 1 = -1? shouldn't it be "1"?
A few definitions:
$$
e^{itheta} = cos(theta) + i sin(theta)
$$
$$
e^{-itheta} = cos(theta) - i sin(theta)
$$
$$
i*i = -1
$$
$$
frac{1}{i} = -i
$$
complex-numbers
complex-numbers
edited Jan 13 at 2:33
Blue
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
edited Jan 13 at 2:33
Blue
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
edited Jan 13 at 2:33
Blue
49.7k870158
edited Jan 13 at 2:33
Blue
49.7k870158
edited Jan 13 at 2:33
Blue
49.7k870158
49.7k870158
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
asked Jan 13 at 1:25
MrCasualityMrCasuality
82
82
7
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
3
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
3
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
2
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
2
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31
|
show 4 more comments
7
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
3
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
3
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
2
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
2
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31
7
7
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
3
3
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
3
3
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
2
2
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
2
2
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The $n$th roots of unity are the roots of the polynomial $x^n - 1$.
If you factor this polynomial as
$$x^n - 1 = prod_{j=0}^{n-1} (x-e^{2pi i j/n}) ,$$
then you will notice by multiplying out and comparing constant terms that the product of the roots is $pm 1$, with the sign depending on whether $n$ is even or odd.
answered Jan 13 at 2:16
DaneDane
3,2491836
$endgroup$
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
add a comment |
$begingroup$
Clearly the original claim is false.
Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
$endgroup$
add a comment |
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2 Answers
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$begingroup$
The $n$th roots of unity are the roots of the polynomial $x^n - 1$.
If you factor this polynomial as
$$x^n - 1 = prod_{j=0}^{n-1} (x-e^{2pi i j/n}) ,$$
then you will notice by multiplying out and comparing constant terms that the product of the roots is $pm 1$, with the sign depending on whether $n$ is even or odd.
answered Jan 13 at 2:16
DaneDane
3,2491836
$endgroup$
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
add a comment |
$begingroup$
The $n$th roots of unity are the roots of the polynomial $x^n - 1$.
If you factor this polynomial as
$$x^n - 1 = prod_{j=0}^{n-1} (x-e^{2pi i j/n}) ,$$
then you will notice by multiplying out and comparing constant terms that the product of the roots is $pm 1$, with the sign depending on whether $n$ is even or odd.
answered Jan 13 at 2:16
DaneDane
3,2491836
$endgroup$
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
add a comment |
$begingroup$
The $n$th roots of unity are the roots of the polynomial $x^n - 1$.
If you factor this polynomial as
$$x^n - 1 = prod_{j=0}^{n-1} (x-e^{2pi i j/n}) ,$$
then you will notice by multiplying out and comparing constant terms that the product of the roots is $pm 1$, with the sign depending on whether $n$ is even or odd.
answered Jan 13 at 2:16
DaneDane
3,2491836
$endgroup$
The $n$th roots of unity are the roots of the polynomial $x^n - 1$.
If you factor this polynomial as
$$x^n - 1 = prod_{j=0}^{n-1} (x-e^{2pi i j/n}) ,$$
then you will notice by multiplying out and comparing constant terms that the product of the roots is $pm 1$, with the sign depending on whether $n$ is even or odd.
answered Jan 13 at 2:16
DaneDane
3,2491836
answered Jan 13 at 2:16
DaneDane
3,2491836
answered Jan 13 at 2:16
DaneDane
3,2491836
answered Jan 13 at 2:16
DaneDane
3,2491836
3,2491836
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
add a comment |
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
3
3
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
$begingroup$
For those that don't know, this is precisely the implication of Vieta's Formula for the constant term.
$endgroup$
– Blue
Jan 13 at 2:27
add a comment |
$begingroup$
Clearly the original claim is false.
Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
$endgroup$
add a comment |
$begingroup$
Clearly the original claim is false.
Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
$endgroup$
add a comment |
$begingroup$
Clearly the original claim is false.
Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
$endgroup$
Clearly the original claim is false.
Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
answered Jan 13 at 2:52
Oscar LanziOscar Lanzi
13.7k12136
13.7k12136
add a comment |
add a comment |
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7
$begingroup$
You seem to have determined that the initial "equality" is sometimes false. Indeed, the product of the $n$-th roots of unity isn't always unity.
$endgroup$
– Blue
Jan 13 at 1:33
3
$begingroup$
"I'm just finding all the nth roots of unity and multiplying them together" ... That's as good a way to spend an evening as any, but ... Who told you that you'd get $1$ as the result? You've shown for yourself that this simply isn't true, even for $n=2$. To ensure that you get $1$ from an $n$-th root of unity, you take the product of $n$ copies of that root.
$endgroup$
– Blue
Jan 13 at 2:01
3
$begingroup$
In $sqrt{2}cdot sqrt{2}=2$ (or $(-sqrt{2})cdot(-sqrt{2})=2$), you're multiplying a particular square root of $2$ by itself, getting $2$; that is what square roots do. But if you multiply the two different square roots of two together, you get $sqrt{2}cdot(-sqrt{2})=-2 neq 2$.
$endgroup$
– Blue
Jan 13 at 2:05
2
$begingroup$
I don't have time to watch a six-minute video to find the source of your confusion. Can you provide a time stamp?
$endgroup$
– Blue
Jan 13 at 2:13
2
$begingroup$
(You either meant $x^2-1$, or else you omitted some $sqrt{2}$s in your factors, but be that as it may ...) I'm happy that someone got through to you. :)
$endgroup$
– Blue
Jan 13 at 2:31