In $mathbb C$ does $sqrt z=z^{1/2}$?












2












$begingroup$


First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?






complex-analysis functions







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    2












    $begingroup$


    First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?






    complex-analysis functions







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?






      complex-analysis functions







      share|cite|improve this question











      $endgroup$




      First of all, is $zmapsto sqrt{z}$ really a function? Indeed, each complex number has two square roots, so I have the impression that $sqrt z$ is not well-defined as a function, is it? Then, I was wondering if $sqrt z=z^{frac{1}{2}}$ was still true. Indeed, $z^{1/2}:=e^{frac{1}{2}log(z)}$, and this is holomorphic in some domain. Whereas, I even have doubt if $zmapsto sqrt z$ is really a function. What do you think ?






      complex-analysis functions




      complex-analysis functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 12:00









      Gaby Alfonso

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      736315










      asked Dec 10 '18 at 11:40









      user623855user623855

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          $begingroup$

          The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.



          Note that the two choices of principal branches agree so that indeed
          $$
          sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
          $$






          share|cite|improve this answer











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            $begingroup$

            The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.



            Note that the two choices of principal branches agree so that indeed
            $$
            sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
            $$






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.



              Note that the two choices of principal branches agree so that indeed
              $$
              sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
              $$






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.



                Note that the two choices of principal branches agree so that indeed
                $$
                sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
                $$






                share|cite|improve this answer











                $endgroup$



                The notations $sqrt{z}$ and $z^{1/2}$ both refer to the principal square root of $z$, which is defined as the solution of $w^2=z$ with argument in $(-pi/2,pi/2]$. Using $log(z)$ you run into the same problem because $exp(w)=z$ even has infinitely many solutions $winBbb C$. This is why for complex numbers you would write $operatorname{Log}(z)$ to refer to the principal value of the logarithm, which is the solution of $exp(w)=z$ with imaginary part in $(-pi,pi]$.



                Note that the two choices of principal branches agree so that indeed
                $$
                sqrt{z} = expleft(frac12 operatorname{Log}(z)right) = z^{1/2}.
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 10 '18 at 12:03

























                answered Dec 10 '18 at 11:45









                ChristophChristoph

                11.9k1642




                11.9k1642






























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