On complex powers of complex numbers












2












$begingroup$


For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
begin{equation}
(-z)^s = e^{-pi is}z^s,
end{equation} but
begin{equation}
(-1)^sz^s = e^{pi i s}z^s
end{equation} and both are not equal.






complex-analysis complex-numbers modular-forms







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
    begin{equation}
    (-z)^s = e^{-pi is}z^s,
    end{equation}     but
    begin{equation}
    (-1)^sz^s = e^{pi i s}z^s
    end{equation}     and both are not equal.






    complex-analysis complex-numbers modular-forms







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
      begin{equation}
      (-z)^s = e^{-pi is}z^s,
      end{equation}       but
      begin{equation}
      (-1)^sz^s = e^{pi i s}z^s
      end{equation}       and both are not equal.






      complex-analysis complex-numbers modular-forms







      share|cite|improve this question











      $endgroup$




      For $z,s in mathbb{C}$ and $zneq 0$, set $z^s = exp(s,log z)$ and $-pi < arg z leq pi$. In this setting, I am worried about the cases where I have to be careful in assuming $(z_1z_2)^s = (z_1)^s(z_2)^s$. As long as $s in mathbb{Z}$, I am safe, but what about the rest? For example, if $zin mathbb{H}$, then
      begin{equation}
      (-z)^s = e^{-pi is}z^s,
      end{equation}       but
      begin{equation}
      (-1)^sz^s = e^{pi i s}z^s
      end{equation}       and both are not equal.






      complex-analysis complex-numbers modular-forms




      complex-analysis complex-numbers modular-forms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 4:18









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 17 '18 at 4:14









      user166305user166305

      1968




      1968






















          1 Answer
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          $begingroup$

          The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
          begin{align}
          (z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
          &=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
          end{align}
          where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
          $$
          z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
          $$
          The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.



          In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.






          share|cite|improve this answer









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            active

            oldest

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            1












            $begingroup$

            The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
            begin{align}
            (z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
            &=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
            end{align}
            where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
            $$
            z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
            $$
            The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.



            In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
              begin{align}
              (z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
              &=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
              end{align}
              where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
              $$
              z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
              $$
              The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.



              In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
                begin{align}
                (z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
                &=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
                end{align}
                where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
                $$
                z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
                $$
                The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.



                In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.






                share|cite|improve this answer









                $endgroup$



                The rule $(zw)^s=z^sw^s$ depends on the equality $log(zw)=log z+log w$. But you need the logarithm to be single valued, which requires limiting the argument to one interval, for instance $(-pi,pi]$ as you did; and this generates the following problem. If $t+u>pi$, then $log (e^{i(t+u)})=i(t+u-2pi)$. Thus when $s=a+ib$, $z_j=r_je^{it_j}$,
                begin{align}
                (z_1z_2)^s&=exp((a+ib)(log r_1r_2+ i(t_1+t_2-2kpi))\ \
                &=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2},e^{2bkpi},
                end{align}
                where $k=0$ if $t_1+t_2leqpi$, and $k=1$ when $t_1+t_2>pi$. On the other hand you have
                $$
                z_1^sz_2^s=r_1^ar_2^a,e^{iblog r_1},e^{iblog r_2},e^{iat_1},e^{iat_2},e^{-bt_1},e^{-bt_2}.
                $$
                The two expressions are equal when $t+uleqpi$, and differ by a factor of $e^{2bpi}$ when $t+u>pi$.



                In summary, $(z_1z_2)^s=z_1^s+z_2^s$ (in your setup) when $arg z_1+arg z_2leqpi$. This is not terrible, but it prevents you from using the formula in general, when you don't know the arguments of $z_1$ and $z_2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 4:51









                Martin ArgeramiMartin Argerami

                125k1178178




                125k1178178






























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