Cube root of a complex number












0












$begingroup$


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?










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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
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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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














  • 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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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. Also right)^{!2} would push the exponent towards the parenthesis.
    $endgroup$
    – egreg
    Aug 17 '18 at 8:56





















2












$begingroup$

Consider equation:



$x^3-1=0$



$$x^3-1=(x-1)(x^2+x+1)=0$$



$$x^2+x+1=0$$



$$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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 $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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. Also right)^{!2} would push the exponent towards the parenthesis.
      $endgroup$
      – egreg
      Aug 17 '18 at 8:56


















    4












    $begingroup$

    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 $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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. Also right)^{!2} would push the exponent towards the parenthesis.
      $endgroup$
      – egreg
      Aug 17 '18 at 8:56
















    4












    4








    4





    $begingroup$

    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 $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$






    share|cite|improve this answer











    $endgroup$



    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 $$exp left(frac{2pi i}3 right)^{!2}=exp left(frac{4pi i}3 right)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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. Also right)^{!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










    • $begingroup$
      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. Also right)^{!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













    2












    $begingroup$

    Consider equation:



    $x^3-1=0$



    $$x^3-1=(x-1)(x^2+x+1)=0$$



    $$x^2+x+1=0$$



    $$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Consider equation:



      $x^3-1=0$



      $$x^3-1=(x-1)(x^2+x+1)=0$$



      $$x^2+x+1=0$$



      $$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Consider equation:



        $x^3-1=0$



        $$x^3-1=(x-1)(x^2+x+1)=0$$



        $$x^2+x+1=0$$



        $$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$






        share|cite|improve this answer









        $endgroup$



        Consider equation:



        $x^3-1=0$



        $$x^3-1=(x-1)(x^2+x+1)=0$$



        $$x^2+x+1=0$$



        $$x=frac{-1±sqrt {1-4}}{2}=frac{-1±sqrt 3 i}{2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 17 '18 at 8:46









        siroussirous

        1,6891514




        1,6891514






























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