Prove that $sin(z) =frac{e^{iz}-e^{-iz}}{2i} =sin(x)cosh(y) +icos(x)sinh(y)$












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For complex $z=x+iy$.
How to prove that
$$sin(z) =frac{e^{iz}-e^{-iz}}{2i} =sin(x)cosh(y) +icos(x)sinh(y)$$
by using the power series definition
$$sin(z)=z-frac{z^3}{3!}+frac{z^5}{5!}-ldots$$
and Euler's formula $$e^z =e^x (cos (y) +isin(y))?$$






complex-analysis algebra-precalculus trigonometry complex-numbers hyperbolic-functions







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    For complex $z=x+iy$.
    How to prove that
    $$sin(z) =frac{e^{iz}-e^{-iz}}{2i} =sin(x)cosh(y) +icos(x)sinh(y)$$
    by using the power series definition
    $$sin(z)=z-frac{z^3}{3!}+frac{z^5}{5!}-ldots$$
    and Euler's formula $$e^z =e^x (cos (y) +isin(y))?$$






    complex-analysis algebra-precalculus trigonometry complex-numbers hyperbolic-functions







    share|cite|improve this question



























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      For complex $z=x+iy$.
      How to prove that
      $$sin(z) =frac{e^{iz}-e^{-iz}}{2i} =sin(x)cosh(y) +icos(x)sinh(y)$$
      by using the power series definition
      $$sin(z)=z-frac{z^3}{3!}+frac{z^5}{5!}-ldots$$
      and Euler's formula $$e^z =e^x (cos (y) +isin(y))?$$






      complex-analysis algebra-precalculus trigonometry complex-numbers hyperbolic-functions







      share|cite|improve this question















      For complex $z=x+iy$.
      How to prove that
      $$sin(z) =frac{e^{iz}-e^{-iz}}{2i} =sin(x)cosh(y) +icos(x)sinh(y)$$
      by using the power series definition
      $$sin(z)=z-frac{z^3}{3!}+frac{z^5}{5!}-ldots$$
      and Euler's formula $$e^z =e^x (cos (y) +isin(y))?$$






      complex-analysis algebra-precalculus trigonometry complex-numbers hyperbolic-functions




      complex-analysis algebra-precalculus trigonometry complex-numbers hyperbolic-functions






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      edited Nov 27 at 12:17









      Martin Sleziak

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      asked Apr 10 at 4:41







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          Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
          begin{align}
          sin z &=sum_{n=0}^infty frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =sum_{n=0}^infty frac{i^{2n}}{(2n+1)!} z^{2n+1}\
          &=frac{1}{i} sum_{n=0}^infty frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =frac{1}{i} sum_{n=0}^infty frac{(iz)^{2n+1}}{(2n+1)!}.
          end{align}
          Consider $dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.



          begin{align}
          sin z &=frac{1}{i} sum_{k=0}^infty left(frac{1-(-1)^k}{2}right) frac{(iz)^k}{k!}
          =frac{1}{2i} sum_{k=0}^infty frac{1-(-1)^k}{k!} (iz)^k\
          &=frac{1}{2i} sum_{k=0}^infty frac{(iz)^k}{k!} -frac{1}{2i} sum_{k=0}^infty frac{(-iz)^k}{k!} \
          &=frac{e^{zi} -e^{-zi}}{2i}\
          &=frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\
          &=frac{e^{-y+xi} -e^{y-xi}}{2i}\
          &=frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\
          &=frac{e^{-y} (cos x +isin x) -e^{y} (cos x -isin x)}{2i}\
          &=frac{i}{i} cdot sin x cdot frac{e^{y} +e^{-y}}{2} -frac{1}{i} cdot cos x cdot frac{e^{y} -e^{-y}}{2}\
          &=sin x cosh y +i cos x sinh y.
          end{align}






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            Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
            begin{align}
            sin z &=sum_{n=0}^infty frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =sum_{n=0}^infty frac{i^{2n}}{(2n+1)!} z^{2n+1}\
            &=frac{1}{i} sum_{n=0}^infty frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =frac{1}{i} sum_{n=0}^infty frac{(iz)^{2n+1}}{(2n+1)!}.
            end{align}
            Consider $dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.



            begin{align}
            sin z &=frac{1}{i} sum_{k=0}^infty left(frac{1-(-1)^k}{2}right) frac{(iz)^k}{k!}
            =frac{1}{2i} sum_{k=0}^infty frac{1-(-1)^k}{k!} (iz)^k\
            &=frac{1}{2i} sum_{k=0}^infty frac{(iz)^k}{k!} -frac{1}{2i} sum_{k=0}^infty frac{(-iz)^k}{k!} \
            &=frac{e^{zi} -e^{-zi}}{2i}\
            &=frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\
            &=frac{e^{-y+xi} -e^{y-xi}}{2i}\
            &=frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\
            &=frac{e^{-y} (cos x +isin x) -e^{y} (cos x -isin x)}{2i}\
            &=frac{i}{i} cdot sin x cdot frac{e^{y} +e^{-y}}{2} -frac{1}{i} cdot cos x cdot frac{e^{y} -e^{-y}}{2}\
            &=sin x cosh y +i cos x sinh y.
            end{align}






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              Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
              begin{align}
              sin z &=sum_{n=0}^infty frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =sum_{n=0}^infty frac{i^{2n}}{(2n+1)!} z^{2n+1}\
              &=frac{1}{i} sum_{n=0}^infty frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =frac{1}{i} sum_{n=0}^infty frac{(iz)^{2n+1}}{(2n+1)!}.
              end{align}
              Consider $dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.



              begin{align}
              sin z &=frac{1}{i} sum_{k=0}^infty left(frac{1-(-1)^k}{2}right) frac{(iz)^k}{k!}
              =frac{1}{2i} sum_{k=0}^infty frac{1-(-1)^k}{k!} (iz)^k\
              &=frac{1}{2i} sum_{k=0}^infty frac{(iz)^k}{k!} -frac{1}{2i} sum_{k=0}^infty frac{(-iz)^k}{k!} \
              &=frac{e^{zi} -e^{-zi}}{2i}\
              &=frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\
              &=frac{e^{-y+xi} -e^{y-xi}}{2i}\
              &=frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\
              &=frac{e^{-y} (cos x +isin x) -e^{y} (cos x -isin x)}{2i}\
              &=frac{i}{i} cdot sin x cdot frac{e^{y} +e^{-y}}{2} -frac{1}{i} cdot cos x cdot frac{e^{y} -e^{-y}}{2}\
              &=sin x cosh y +i cos x sinh y.
              end{align}






              share|cite|improve this answer


























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                Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
                begin{align}
                sin z &=sum_{n=0}^infty frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =sum_{n=0}^infty frac{i^{2n}}{(2n+1)!} z^{2n+1}\
                &=frac{1}{i} sum_{n=0}^infty frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =frac{1}{i} sum_{n=0}^infty frac{(iz)^{2n+1}}{(2n+1)!}.
                end{align}
                Consider $dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.



                begin{align}
                sin z &=frac{1}{i} sum_{k=0}^infty left(frac{1-(-1)^k}{2}right) frac{(iz)^k}{k!}
                =frac{1}{2i} sum_{k=0}^infty frac{1-(-1)^k}{k!} (iz)^k\
                &=frac{1}{2i} sum_{k=0}^infty frac{(iz)^k}{k!} -frac{1}{2i} sum_{k=0}^infty frac{(-iz)^k}{k!} \
                &=frac{e^{zi} -e^{-zi}}{2i}\
                &=frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\
                &=frac{e^{-y+xi} -e^{y-xi}}{2i}\
                &=frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\
                &=frac{e^{-y} (cos x +isin x) -e^{y} (cos x -isin x)}{2i}\
                &=frac{i}{i} cdot sin x cdot frac{e^{y} +e^{-y}}{2} -frac{1}{i} cdot cos x cdot frac{e^{y} -e^{-y}}{2}\
                &=sin x cosh y +i cos x sinh y.
                end{align}






                share|cite|improve this answer














                Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
                begin{align}
                sin z &=sum_{n=0}^infty frac{(-1)^{n}}{(2n+1)!} z^{2n+1} =sum_{n=0}^infty frac{i^{2n}}{(2n+1)!} z^{2n+1}\
                &=frac{1}{i} sum_{n=0}^infty frac{i^{2n+1}}{(2n+1)!} z^{2n+1} =frac{1}{i} sum_{n=0}^infty frac{(iz)^{2n+1}}{(2n+1)!}.
                end{align}
                Consider $dfrac{1-(-1)^k}{2}$ which equals $1$ for odd $k$ and $0$ for even $k$.



                begin{align}
                sin z &=frac{1}{i} sum_{k=0}^infty left(frac{1-(-1)^k}{2}right) frac{(iz)^k}{k!}
                =frac{1}{2i} sum_{k=0}^infty frac{1-(-1)^k}{k!} (iz)^k\
                &=frac{1}{2i} sum_{k=0}^infty frac{(iz)^k}{k!} -frac{1}{2i} sum_{k=0}^infty frac{(-iz)^k}{k!} \
                &=frac{e^{zi} -e^{-zi}}{2i}\
                &=frac{e^{(x+yi)i} -e^{-(x+yi)i}}{2i}\
                &=frac{e^{-y+xi} -e^{y-xi}}{2i}\
                &=frac{e^{-y} e^{xi} -e^{y} e^{-xi}}{2i}\
                &=frac{e^{-y} (cos x +isin x) -e^{y} (cos x -isin x)}{2i}\
                &=frac{i}{i} cdot sin x cdot frac{e^{y} +e^{-y}}{2} -frac{1}{i} cdot cos x cdot frac{e^{y} -e^{-y}}{2}\
                &=sin x cosh y +i cos x sinh y.
                end{align}







                share|cite|improve this answer














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                edited Nov 18 at 18:00

























                answered Apr 10 at 10:08









                Rócherz

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