If $z=costheta + isin theta$ prove $frac{z^2-1}{z^2+1}=itantheta$












3












$begingroup$



If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$




Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.



$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$



I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.






trigonometry complex-numbers







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$endgroup$












  • $begingroup$
    Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
    $endgroup$
    – Doug M
    Jan 14 at 21:58


















3












$begingroup$



If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$




Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.



$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$



I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.






trigonometry complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
    $endgroup$
    – Doug M
    Jan 14 at 21:58
















3












3








3





$begingroup$



If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$




Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.



$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$



I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.






trigonometry complex-numbers







share|cite|improve this question











$endgroup$





If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$




Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.



$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$



I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.






trigonometry complex-numbers




trigonometry complex-numbers






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edited Jan 14 at 21:48







H.Linkhorn

















asked Jan 14 at 21:29









H.LinkhornH.Linkhorn

383113




383113












  • $begingroup$
    Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
    $endgroup$
    – Doug M
    Jan 14 at 21:58




















  • $begingroup$
    Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
    $endgroup$
    – Doug M
    Jan 14 at 21:58


















$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58






$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58












5 Answers
5






active

oldest

votes


















4












$begingroup$

Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    $z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $cos theta +isin theta =e ^{i theta}$. Then



      $$begin{aligned}
      frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
      &=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
      &=frac{2 i sin theta}{2 cos theta}\
      &=itantheta
      end{aligned}$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        If you divide the numerator and denominator of



        $$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$



        by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain



        $$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$



        which simplifies to



        $$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$



        Then factor the numerator to get



        $$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$



        Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          Now where you've left off,



          use $cos2y=2cos^2y-1=1-2sin^2y (1)$



          to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$



          Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
          use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$






          share|cite|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.






                share|cite|improve this answer









                $endgroup$



                Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 21:33









                J.G.J.G.

                25.5k22539




                25.5k22539























                    3












                    $begingroup$

                    $z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$






                    share|cite|improve this answer









                    $endgroup$


















                      3












                      $begingroup$

                      $z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$






                      share|cite|improve this answer









                      $endgroup$
















                        3












                        3








                        3





                        $begingroup$

                        $z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$






                        share|cite|improve this answer









                        $endgroup$



                        $z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 14 at 21:38









                        MahranMahran

                        634




                        634























                            0












                            $begingroup$

                            $cos theta +isin theta =e ^{i theta}$. Then



                            $$begin{aligned}
                            frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
                            &=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
                            &=frac{2 i sin theta}{2 cos theta}\
                            &=itantheta
                            end{aligned}$$






                            share|cite|improve this answer









                            $endgroup$


















                              0












                              $begingroup$

                              $cos theta +isin theta =e ^{i theta}$. Then



                              $$begin{aligned}
                              frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
                              &=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
                              &=frac{2 i sin theta}{2 cos theta}\
                              &=itantheta
                              end{aligned}$$






                              share|cite|improve this answer









                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                $cos theta +isin theta =e ^{i theta}$. Then



                                $$begin{aligned}
                                frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
                                &=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
                                &=frac{2 i sin theta}{2 cos theta}\
                                &=itantheta
                                end{aligned}$$






                                share|cite|improve this answer









                                $endgroup$



                                $cos theta +isin theta =e ^{i theta}$. Then



                                $$begin{aligned}
                                frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
                                &=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
                                &=frac{2 i sin theta}{2 cos theta}\
                                &=itantheta
                                end{aligned}$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 14 at 21:37









                                mathcounterexamples.netmathcounterexamples.net

                                26.5k22157




                                26.5k22157























                                    0












                                    $begingroup$

                                    If you divide the numerator and denominator of



                                    $$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$



                                    by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain



                                    $$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$



                                    which simplifies to



                                    $$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$



                                    Then factor the numerator to get



                                    $$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$



                                    Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      If you divide the numerator and denominator of



                                      $$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$



                                      by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain



                                      $$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$



                                      which simplifies to



                                      $$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$



                                      Then factor the numerator to get



                                      $$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$



                                      Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        If you divide the numerator and denominator of



                                        $$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$



                                        by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain



                                        $$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$



                                        which simplifies to



                                        $$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$



                                        Then factor the numerator to get



                                        $$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$



                                        Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$






                                        share|cite|improve this answer









                                        $endgroup$



                                        If you divide the numerator and denominator of



                                        $$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$



                                        by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain



                                        $$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$



                                        which simplifies to



                                        $$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$



                                        Then factor the numerator to get



                                        $$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$



                                        Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jan 14 at 22:07









                                        John Wayland BalesJohn Wayland Bales

                                        14k21238




                                        14k21238























                                            0












                                            $begingroup$

                                            Now where you've left off,



                                            use $cos2y=2cos^2y-1=1-2sin^2y (1)$



                                            to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$



                                            Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
                                            use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Now where you've left off,



                                              use $cos2y=2cos^2y-1=1-2sin^2y (1)$



                                              to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$



                                              Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
                                              use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Now where you've left off,



                                                use $cos2y=2cos^2y-1=1-2sin^2y (1)$



                                                to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$



                                                Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
                                                use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$






                                                share|cite|improve this answer









                                                $endgroup$



                                                Now where you've left off,



                                                use $cos2y=2cos^2y-1=1-2sin^2y (1)$



                                                to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$



                                                Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
                                                use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 15 at 4:25









                                                lab bhattacharjeelab bhattacharjee

                                                225k15157275




                                                225k15157275






























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