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 $(-1 + sqrt{3i})/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt{3i})/2$ squared is $(-1 - sqrt{3i})/2$. 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?
complex-numbers arithmetic
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add a comment |
$begingroup$
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt{3i})/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt{3i})/2$ squared is $(-1 - sqrt{3i})/2$. 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?
complex-numbers arithmetic
$endgroup$
1
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
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– Bernard
Aug 17 '18 at 8:14
3
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54
add a comment |
$begingroup$
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt{3i})/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt{3i})/2$ squared is $(-1 - sqrt{3i})/2$. 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?
complex-numbers arithmetic
$endgroup$
I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:
Show that $(-1 + sqrt{3i})/2$ is a cube root of $1$.
The solution on linearalgebras.com shows the following solution here, at number 2.
It states that $(-1 + sqrt{3i})/2$ squared is $(-1 - sqrt{3i})/2$. 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?
complex-numbers arithmetic
complex-numbers arithmetic
edited Dec 19 '18 at 21:03
Eric Wofsey
187k14215344
187k14215344
asked Aug 17 '18 at 6:30
JaigusJaigus
2259
2259
1
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
$endgroup$
– Bernard
Aug 17 '18 at 8:14
3
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54
add a comment |
1
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
$endgroup$
– Bernard
Aug 17 '18 at 8:14
3
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54
1
1
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
$endgroup$
– Bernard
Aug 17 '18 at 8:14
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
$endgroup$
– Bernard
Aug 17 '18 at 8:14
3
3
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The $i$ should not be inside the square root.
It should be $$left(frac{-1+sqrt3 ,i}{2}right)^{!2}=frac{-1-sqrt3 ,i}{2}$$
rather than
$$left(frac{-1+sqrt{3 i}}{2}right)^{!2}=frac{-1-sqrt{3 i}}{2}$$
It might be easier to understand it as $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$
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$begingroup$
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
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– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
$begingroup$
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)^{!2}
would push the exponent towards the parenthesis.
$endgroup$
– egreg
Aug 17 '18 at 8:56
add a comment |
$begingroup$
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $i$ should not be inside the square root.
It should be $$left(frac{-1+sqrt3 ,i}{2}right)^{!2}=frac{-1-sqrt3 ,i}{2}$$
rather than
$$left(frac{-1+sqrt{3 i}}{2}right)^{!2}=frac{-1-sqrt{3 i}}{2}$$
It might be easier to understand it as $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$
$endgroup$
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
$begingroup$
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)^{!2}
would push the exponent towards the parenthesis.
$endgroup$
– egreg
Aug 17 '18 at 8:56
add a comment |
$begingroup$
The $i$ should not be inside the square root.
It should be $$left(frac{-1+sqrt3 ,i}{2}right)^{!2}=frac{-1-sqrt3 ,i}{2}$$
rather than
$$left(frac{-1+sqrt{3 i}}{2}right)^{!2}=frac{-1-sqrt{3 i}}{2}$$
It might be easier to understand it as $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$
$endgroup$
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
$begingroup$
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)^{!2}
would push the exponent towards the parenthesis.
$endgroup$
– egreg
Aug 17 '18 at 8:56
add a comment |
$begingroup$
The $i$ should not be inside the square root.
It should be $$left(frac{-1+sqrt3 ,i}{2}right)^{!2}=frac{-1-sqrt3 ,i}{2}$$
rather than
$$left(frac{-1+sqrt{3 i}}{2}right)^{!2}=frac{-1-sqrt{3 i}}{2}$$
It might be easier to understand it as $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$
$endgroup$
The $i$ should not be inside the square root.
It should be $$left(frac{-1+sqrt3 ,i}{2}right)^{!2}=frac{-1-sqrt3 ,i}{2}$$
rather than
$$left(frac{-1+sqrt{3 i}}{2}right)^{!2}=frac{-1-sqrt{3 i}}{2}$$
It might be easier to understand it as $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$
edited Aug 18 '18 at 14:57
answered Aug 17 '18 at 6:35
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
$begingroup$
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)^{!2}
would push the exponent towards the parenthesis.
$endgroup$
– egreg
Aug 17 '18 at 8:56
add a comment |
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
$begingroup$
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)^{!2}
would push the exponent towards the parenthesis.
$endgroup$
– egreg
Aug 17 '18 at 8:56
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
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
$endgroup$
– DanielWainfleet
Aug 17 '18 at 8:35
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
$begingroup$
ah, good suggestion.
$endgroup$
– Siong Thye Goh
Aug 17 '18 at 8:37
2
2
$begingroup$
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.$endgroup$
– egreg
Aug 17 '18 at 8:56
$begingroup$
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.$endgroup$
– egreg
Aug 17 '18 at 8:56
add a comment |
$begingroup$
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$
$endgroup$
add a comment |
$begingroup$
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$
$endgroup$
add a comment |
$begingroup$
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$
$endgroup$
Consider equation:
$x^3-1=0$
$$x^3-1=(x-1)(x^2+x+1)=0$$
$$x^2+x+1=0$$
$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$
answered Aug 17 '18 at 8:46
siroussirous
1,6891514
1,6891514
add a comment |
add a comment |
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1
$begingroup$
The symbol $sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number.
$endgroup$
– Bernard
Aug 17 '18 at 8:14
3
$begingroup$
Most certainly it is $sqrt{3},i$, rather than $sqrt{3i}$.
$endgroup$
– egreg
Aug 17 '18 at 8:54