Is it possible to break $f(z)=e^{z^e}$ into real and complex parts?












2












$begingroup$


Example: $h(z)=z^2$
$$h(z)=f(x+iy)=(x+iy)^2$$
$$=x^2-y^2+i(2xy)$$



I am wondering if I could do the same for $f(z)=e^{z^e}$. Any help would be greatly appreciated. Thanks.






complex-analysis complex-numbers







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Example: $h(z)=z^2$
    $$h(z)=f(x+iy)=(x+iy)^2$$
    $$=x^2-y^2+i(2xy)$$



    I am wondering if I could do the same for $f(z)=e^{z^e}$. Any help would be greatly appreciated. Thanks.






    complex-analysis complex-numbers







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Example: $h(z)=z^2$
      $$h(z)=f(x+iy)=(x+iy)^2$$
      $$=x^2-y^2+i(2xy)$$



      I am wondering if I could do the same for $f(z)=e^{z^e}$. Any help would be greatly appreciated. Thanks.






      complex-analysis complex-numbers







      share|cite|improve this question









      $endgroup$




      Example: $h(z)=z^2$
      $$h(z)=f(x+iy)=(x+iy)^2$$
      $$=x^2-y^2+i(2xy)$$



      I am wondering if I could do the same for $f(z)=e^{z^e}$. Any help would be greatly appreciated. Thanks.






      complex-analysis complex-numbers




      complex-analysis complex-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 2 at 12:15









      LokLok

      404




      404






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          First, remember that $w^z = e ^ {z log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
          $$z^e = e^{e log z} = e^{e (log r + i theta)}$$
          where, for the sake of simplicity, I used polar coordinates.



          Remembering Euler's fomula, last expression is equal to $$e^{e log r}(cos{etheta} + i sin{ etheta})$$
          Now, $$e^{z^e}=e ^ {e^{e log r} (cos{e theta} + i sin {e theta})}$$
          and, using again Euler's formula, we get:
          $$e^{e^{e log r} cos{e theta}}bigg(cos{(e^{e log r}sin{etheta}) + i sin(e^{e log r} sin {etheta})bigg)}$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(cosvarphi + isinvarphi)$,
            $$
            z^e = r^e(cos(evarphi+2epi n)+isin(evarphi+2epi n)), qquad ninmathbb Z
            $$



            If you are ok with that, then
            $$
            e^{z^e} = e^{r^ecos(evarphi+2epi n)}Big(cosleft(r^esin(evarphi+2epi n)right)
            +isinleft(r^esin(evarphi+2epi n)right)Big),qquad ninmathbb Z
            $$



            If we want to choose principal branch of $log z$, then $varphi=arg z$ and $n=0$.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Maple says
              $$
              expleft(z^rm{e}right)
              ={{rm e}^{ left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}cos left( {
              rm e}arctan left( y,x right) right) }}cos left( left( {x}^{2
              }+{y}^{2} right) ^{{rm e}/2}sin left( {rm e}arctan left( y,x
              right) right) right)
              \+i{{rm e}^{ left( {x}^{2}+{y}^{2}
              right) ^{{rm e}/2}cos left( {rm e}arctan left( y,x right)
              right) }}sin left( left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}
              sin left( {rm e}arctan left( y,x right) right) right)
              $$






              share|cite|improve this answer









              $endgroup$














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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                First, remember that $w^z = e ^ {z log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
                $$z^e = e^{e log z} = e^{e (log r + i theta)}$$
                where, for the sake of simplicity, I used polar coordinates.



                Remembering Euler's fomula, last expression is equal to $$e^{e log r}(cos{etheta} + i sin{ etheta})$$
                Now, $$e^{z^e}=e ^ {e^{e log r} (cos{e theta} + i sin {e theta})}$$
                and, using again Euler's formula, we get:
                $$e^{e^{e log r} cos{e theta}}bigg(cos{(e^{e log r}sin{etheta}) + i sin(e^{e log r} sin {etheta})bigg)}$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  First, remember that $w^z = e ^ {z log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
                  $$z^e = e^{e log z} = e^{e (log r + i theta)}$$
                  where, for the sake of simplicity, I used polar coordinates.



                  Remembering Euler's fomula, last expression is equal to $$e^{e log r}(cos{etheta} + i sin{ etheta})$$
                  Now, $$e^{z^e}=e ^ {e^{e log r} (cos{e theta} + i sin {e theta})}$$
                  and, using again Euler's formula, we get:
                  $$e^{e^{e log r} cos{e theta}}bigg(cos{(e^{e log r}sin{etheta}) + i sin(e^{e log r} sin {etheta})bigg)}$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    First, remember that $w^z = e ^ {z log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
                    $$z^e = e^{e log z} = e^{e (log r + i theta)}$$
                    where, for the sake of simplicity, I used polar coordinates.



                    Remembering Euler's fomula, last expression is equal to $$e^{e log r}(cos{etheta} + i sin{ etheta})$$
                    Now, $$e^{z^e}=e ^ {e^{e log r} (cos{e theta} + i sin {e theta})}$$
                    and, using again Euler's formula, we get:
                    $$e^{e^{e log r} cos{e theta}}bigg(cos{(e^{e log r}sin{etheta}) + i sin(e^{e log r} sin {etheta})bigg)}$$






                    share|cite|improve this answer









                    $endgroup$



                    First, remember that $w^z = e ^ {z log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
                    $$z^e = e^{e log z} = e^{e (log r + i theta)}$$
                    where, for the sake of simplicity, I used polar coordinates.



                    Remembering Euler's fomula, last expression is equal to $$e^{e log r}(cos{etheta} + i sin{ etheta})$$
                    Now, $$e^{z^e}=e ^ {e^{e log r} (cos{e theta} + i sin {e theta})}$$
                    and, using again Euler's formula, we get:
                    $$e^{e^{e log r} cos{e theta}}bigg(cos{(e^{e log r}sin{etheta}) + i sin(e^{e log r} sin {etheta})bigg)}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 2 at 12:43









                    HarnakHarnak

                    1,369512




                    1,369512























                        2












                        $begingroup$

                        The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(cosvarphi + isinvarphi)$,
                        $$
                        z^e = r^e(cos(evarphi+2epi n)+isin(evarphi+2epi n)), qquad ninmathbb Z
                        $$



                        If you are ok with that, then
                        $$
                        e^{z^e} = e^{r^ecos(evarphi+2epi n)}Big(cosleft(r^esin(evarphi+2epi n)right)
                        +isinleft(r^esin(evarphi+2epi n)right)Big),qquad ninmathbb Z
                        $$



                        If we want to choose principal branch of $log z$, then $varphi=arg z$ and $n=0$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(cosvarphi + isinvarphi)$,
                          $$
                          z^e = r^e(cos(evarphi+2epi n)+isin(evarphi+2epi n)), qquad ninmathbb Z
                          $$



                          If you are ok with that, then
                          $$
                          e^{z^e} = e^{r^ecos(evarphi+2epi n)}Big(cosleft(r^esin(evarphi+2epi n)right)
                          +isinleft(r^esin(evarphi+2epi n)right)Big),qquad ninmathbb Z
                          $$



                          If we want to choose principal branch of $log z$, then $varphi=arg z$ and $n=0$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(cosvarphi + isinvarphi)$,
                            $$
                            z^e = r^e(cos(evarphi+2epi n)+isin(evarphi+2epi n)), qquad ninmathbb Z
                            $$



                            If you are ok with that, then
                            $$
                            e^{z^e} = e^{r^ecos(evarphi+2epi n)}Big(cosleft(r^esin(evarphi+2epi n)right)
                            +isinleft(r^esin(evarphi+2epi n)right)Big),qquad ninmathbb Z
                            $$



                            If we want to choose principal branch of $log z$, then $varphi=arg z$ and $n=0$.






                            share|cite|improve this answer









                            $endgroup$



                            The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(cosvarphi + isinvarphi)$,
                            $$
                            z^e = r^e(cos(evarphi+2epi n)+isin(evarphi+2epi n)), qquad ninmathbb Z
                            $$



                            If you are ok with that, then
                            $$
                            e^{z^e} = e^{r^ecos(evarphi+2epi n)}Big(cosleft(r^esin(evarphi+2epi n)right)
                            +isinleft(r^esin(evarphi+2epi n)right)Big),qquad ninmathbb Z
                            $$



                            If we want to choose principal branch of $log z$, then $varphi=arg z$ and $n=0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 2 at 12:35









                            Vasily MitchVasily Mitch

                            2,6791312




                            2,6791312























                                1












                                $begingroup$

                                Maple says
                                $$
                                expleft(z^rm{e}right)
                                ={{rm e}^{ left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}cos left( {
                                rm e}arctan left( y,x right) right) }}cos left( left( {x}^{2
                                }+{y}^{2} right) ^{{rm e}/2}sin left( {rm e}arctan left( y,x
                                right) right) right)
                                \+i{{rm e}^{ left( {x}^{2}+{y}^{2}
                                right) ^{{rm e}/2}cos left( {rm e}arctan left( y,x right)
                                right) }}sin left( left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}
                                sin left( {rm e}arctan left( y,x right) right) right)
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Maple says
                                  $$
                                  expleft(z^rm{e}right)
                                  ={{rm e}^{ left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}cos left( {
                                  rm e}arctan left( y,x right) right) }}cos left( left( {x}^{2
                                  }+{y}^{2} right) ^{{rm e}/2}sin left( {rm e}arctan left( y,x
                                  right) right) right)
                                  \+i{{rm e}^{ left( {x}^{2}+{y}^{2}
                                  right) ^{{rm e}/2}cos left( {rm e}arctan left( y,x right)
                                  right) }}sin left( left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}
                                  sin left( {rm e}arctan left( y,x right) right) right)
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Maple says
                                    $$
                                    expleft(z^rm{e}right)
                                    ={{rm e}^{ left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}cos left( {
                                    rm e}arctan left( y,x right) right) }}cos left( left( {x}^{2
                                    }+{y}^{2} right) ^{{rm e}/2}sin left( {rm e}arctan left( y,x
                                    right) right) right)
                                    \+i{{rm e}^{ left( {x}^{2}+{y}^{2}
                                    right) ^{{rm e}/2}cos left( {rm e}arctan left( y,x right)
                                    right) }}sin left( left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}
                                    sin left( {rm e}arctan left( y,x right) right) right)
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Maple says
                                    $$
                                    expleft(z^rm{e}right)
                                    ={{rm e}^{ left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}cos left( {
                                    rm e}arctan left( y,x right) right) }}cos left( left( {x}^{2
                                    }+{y}^{2} right) ^{{rm e}/2}sin left( {rm e}arctan left( y,x
                                    right) right) right)
                                    \+i{{rm e}^{ left( {x}^{2}+{y}^{2}
                                    right) ^{{rm e}/2}cos left( {rm e}arctan left( y,x right)
                                    right) }}sin left( left( {x}^{2}+{y}^{2} right) ^{{rm e}/2}
                                    sin left( {rm e}arctan left( y,x right) right) right)
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Feb 2 at 12:45









                                    GEdgarGEdgar

                                    63.5k269175




                                    63.5k269175






























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