Sum of square roots of complex numbers

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Is this always true?



$z^1/2+z^1/2=2z^1/2$



I said that it isn't in the multiform case but I want to justify this better (not just words).




complex-numbers radicals




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  • 1




    Before using the word “true” in this context, you should define $x^1/2$ first.
    – José Carlos Santos
    Aug 2 at 14:53






  • 1




    If you are sensible enough to take the same square root each time...
    – Lord Shark the Unknown
    Aug 2 at 14:53










  • Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
    – Verónica
    Aug 2 at 15:12










  • @Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
    – Jam
    Aug 2 at 15:14















up vote
1
down vote

favorite












Is this always true?



$z^1/2+z^1/2=2z^1/2$



I said that it isn't in the multiform case but I want to justify this better (not just words).




complex-numbers radicals




share|cite|improve this question

















  • 1




    Before using the word “true” in this context, you should define $x^1/2$ first.
    – José Carlos Santos
    Aug 2 at 14:53






  • 1




    If you are sensible enough to take the same square root each time...
    – Lord Shark the Unknown
    Aug 2 at 14:53










  • Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
    – Verónica
    Aug 2 at 15:12










  • @Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
    – Jam
    Aug 2 at 15:14













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is this always true?



$z^1/2+z^1/2=2z^1/2$



I said that it isn't in the multiform case but I want to justify this better (not just words).




complex-numbers radicals




share|cite|improve this question













Is this always true?



$z^1/2+z^1/2=2z^1/2$



I said that it isn't in the multiform case but I want to justify this better (not just words).





complex-numbers radicals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 14:53









José Carlos Santos

112k1696172




112k1696172









asked Aug 2 at 14:52









Verónica

235




235







  • 1




    Before using the word “true” in this context, you should define $x^1/2$ first.
    – José Carlos Santos
    Aug 2 at 14:53






  • 1




    If you are sensible enough to take the same square root each time...
    – Lord Shark the Unknown
    Aug 2 at 14:53










  • Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
    – Verónica
    Aug 2 at 15:12










  • @Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
    – Jam
    Aug 2 at 15:14













  • 1




    Before using the word “true” in this context, you should define $x^1/2$ first.
    – José Carlos Santos
    Aug 2 at 14:53






  • 1




    If you are sensible enough to take the same square root each time...
    – Lord Shark the Unknown
    Aug 2 at 14:53










  • Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
    – Verónica
    Aug 2 at 15:12










  • @Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
    – Jam
    Aug 2 at 15:14








1




1




Before using the word “true” in this context, you should define $x^1/2$ first.
– José Carlos Santos
Aug 2 at 14:53




Before using the word “true” in this context, you should define $x^1/2$ first.
– José Carlos Santos
Aug 2 at 14:53




1




1




If you are sensible enough to take the same square root each time...
– Lord Shark the Unknown
Aug 2 at 14:53




If you are sensible enough to take the same square root each time...
– Lord Shark the Unknown
Aug 2 at 14:53












Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
– Verónica
Aug 2 at 15:12




Why's necessary? It's true if you define the square root but I'mt trying to say that is false if you don't define it.
– Verónica
Aug 2 at 15:12












@Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
– Jam
Aug 2 at 15:14





@Verónica The square root function can be a bit ambiguous since you could be talking about $+sqrtz$ or $-sqrtz$, but this is confounded even more since in the complex plane, $a^b=exp(bln(a)+2pi n)$. So without defining exactly what you mean (or exactly what branch cut you mean), there's no way of knowing whether your equation holds.
– Jam
Aug 2 at 15:14











1 Answer
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For $zneq 0$ we have $2$ distinct solutions



$$z^frac12=wimplies w_1^2=zquad w_2^2=z$$



and for any solution of course



$$w_i+w_i=2w_i$$



but



$$w_1+w_2=0$$






share|cite|improve this answer





















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    up vote
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    For $zneq 0$ we have $2$ distinct solutions



    $$z^frac12=wimplies w_1^2=zquad w_2^2=z$$



    and for any solution of course



    $$w_i+w_i=2w_i$$



    but



    $$w_1+w_2=0$$






    share|cite|improve this answer

























      up vote
      1
      down vote













      For $zneq 0$ we have $2$ distinct solutions



      $$z^frac12=wimplies w_1^2=zquad w_2^2=z$$



      and for any solution of course



      $$w_i+w_i=2w_i$$



      but



      $$w_1+w_2=0$$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        For $zneq 0$ we have $2$ distinct solutions



        $$z^frac12=wimplies w_1^2=zquad w_2^2=z$$



        and for any solution of course



        $$w_i+w_i=2w_i$$



        but



        $$w_1+w_2=0$$






        share|cite|improve this answer













        For $zneq 0$ we have $2$ distinct solutions



        $$z^frac12=wimplies w_1^2=zquad w_2^2=z$$



        and for any solution of course



        $$w_i+w_i=2w_i$$



        but



        $$w_1+w_2=0$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 15:12









        gimusi

        63.8k73480




        63.8k73480






















             

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