Cube root of a complex number
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I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt3i)/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt3i)/2$ squared is $(-1 - sqrt3i)/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?
linear-algebra discrete-mathematics complex-numbers proof-writing
edited Aug 17 at 8:12
Bernard
111k635103
asked Aug 17 at 6:30
Jaigus
1032
add a comment |Â
up vote
0
down vote
favorite
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt3i)/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt3i)/2$ squared is $(-1 - sqrt3i)/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?
linear-algebra discrete-mathematics complex-numbers proof-writing
edited Aug 17 at 8:12
Bernard
111k635103
asked Aug 17 at 6:30
Jaigus
1032
1
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
2
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt3i)/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt3i)/2$ squared is $(-1 - sqrt3i)/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?
linear-algebra discrete-mathematics complex-numbers proof-writing
edited Aug 17 at 8:12
Bernard
111k635103
asked Aug 17 at 6:30
Jaigus
1032
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt3i)/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt3i)/2$ squared is $(-1 - sqrt3i)/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?
linear-algebra discrete-mathematics complex-numbers proof-writing
edited Aug 17 at 8:12
Bernard
111k635103
asked Aug 17 at 6:30
Jaigus
1032
edited Aug 17 at 8:12
Bernard
111k635103
edited Aug 17 at 8:12
Bernard
111k635103
edited Aug 17 at 8:12
Bernard
111k635103
111k635103
asked Aug 17 at 6:30
Jaigus
1032
asked Aug 17 at 6:30
Jaigus
1032
asked Aug 17 at 6:30
Jaigus
1032
1032
1
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
2
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54
add a comment |Â
1
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
2
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54
1
1
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
2
2
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The $i$ should not be inside the square root.
It should be $$left(frac-1+sqrt3 ,i2right)^!2=frac-1-sqrt3 ,i2$$
rather than
$$left(frac-1+sqrt3 i2right)^!2=frac-1-sqrt3 i2$$
It might be easier to understand it as $$exp left(frac2pi i3 right)^!2=exp left(frac4pi i3 right)$$
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
Better,than;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space,is most likely needed. Alsoright)^!2would push the exponent towards the parenthesis.
– egreg
Aug 17 at 8:56
add a comment |Â
up vote
2
down vote
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac-1±sqrt 1-42=frac-1±sqrt 3 i2$$
answered Aug 17 at 8:46
sirous
876511
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The $i$ should not be inside the square root.
It should be $$left(frac-1+sqrt3 ,i2right)^!2=frac-1-sqrt3 ,i2$$
rather than
$$left(frac-1+sqrt3 i2right)^!2=frac-1-sqrt3 i2$$
It might be easier to understand it as $$exp left(frac2pi i3 right)^!2=exp left(frac4pi i3 right)$$
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
Better,than;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space,is most likely needed. Alsoright)^!2would push the exponent towards the parenthesis.
– egreg
Aug 17 at 8:56
add a comment |Â
up vote
4
down vote
accepted
The $i$ should not be inside the square root.
It should be $$left(frac-1+sqrt3 ,i2right)^!2=frac-1-sqrt3 ,i2$$
rather than
$$left(frac-1+sqrt3 i2right)^!2=frac-1-sqrt3 i2$$
It might be easier to understand it as $$exp left(frac2pi i3 right)^!2=exp left(frac4pi i3 right)$$
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
Better,than;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space,is most likely needed. Alsoright)^!2would push the exponent towards the parenthesis.
– egreg
Aug 17 at 8:56
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The $i$ should not be inside the square root.
It should be $$left(frac-1+sqrt3 ,i2right)^!2=frac-1-sqrt3 ,i2$$
rather than
$$left(frac-1+sqrt3 i2right)^!2=frac-1-sqrt3 i2$$
It might be easier to understand it as $$exp left(frac2pi i3 right)^!2=exp left(frac4pi i3 right)$$
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
The $i$ should not be inside the square root.
It should be $$left(frac-1+sqrt3 ,i2right)^!2=frac-1-sqrt3 ,i2$$
rather than
$$left(frac-1+sqrt3 i2right)^!2=frac-1-sqrt3 i2$$
It might be easier to understand it as $$exp left(frac2pi i3 right)^!2=exp left(frac4pi i3 right)$$
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
edited Aug 18 at 14:57
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
answered Aug 17 at 6:35
Siong Thye Goh
79.7k135299
79.7k135299
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
Better,than;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space,is most likely needed. Alsoright)^!2would push the exponent towards the parenthesis.
– egreg
Aug 17 at 8:56
add a comment |Â
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
Better,than;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space,is most likely needed. Alsoright)^!2would push the exponent towards the parenthesis.
– egreg
Aug 17 at 8:56
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
I often type ; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1
– DanielWainfleet
Aug 17 at 8:35
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
ah, good suggestion.
– Siong Thye Goh
Aug 17 at 8:37
2
2
Better
, than ;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space , is most likely needed. Also right)^!2 would push the exponent towards the parenthesis.– egreg
Aug 17 at 8:56
Better
, than ;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space , is most likely needed. Also right)^!2 would push the exponent towards the parenthesis.– egreg
Aug 17 at 8:56
add a comment |Â
up vote
2
down vote
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac-1±sqrt 1-42=frac-1±sqrt 3 i2$$
answered Aug 17 at 8:46
sirous
876511
add a comment |Â
up vote
2
down vote
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac-1±sqrt 1-42=frac-1±sqrt 3 i2$$
answered Aug 17 at 8:46
sirous
876511
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac-1±sqrt 1-42=frac-1±sqrt 3 i2$$
answered Aug 17 at 8:46
sirous
876511
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac-1±sqrt 1-42=frac-1±sqrt 3 i2$$
answered Aug 17 at 8:46
sirous
876511
answered Aug 17 at 8:46
sirous
876511
answered Aug 17 at 8:46
sirous
876511
answered Aug 17 at 8:46
sirous
876511
876511
add a comment |Â
add a comment |Â
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1
The symbol $sqrt~$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
– Bernard
Aug 17 at 8:14
2
Most certainly it is $sqrt3,i$, rather than $sqrt3i$.
– egreg
Aug 17 at 8:54