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












1














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





























    1














    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



























      1












      1








      1


      1





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 at 12:17









      Martin Sleziak

      44.7k7115270




      44.7k7115270










      asked Apr 10 at 4:41







      user546987





























          1 Answer
          1






          active

          oldest

          votes


















          0














          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2730426%2fprove-that-sinz-fraceiz-e-iz2i-sinx-coshy-i-cosx-sinhy%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown
























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            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




























              0














              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


























                0












                0








                0






                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 18 at 18:00

























                answered Apr 10 at 10:08









                Rócherz

                2,7762721




                2,7762721






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2730426%2fprove-that-sinz-fraceiz-e-iz2i-sinx-coshy-i-cosx-sinhy%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more