Differentiating polar functions using complex numbers












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I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?



For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?






derivatives complex-numbers polar-coordinates







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  • $begingroup$
    en.wikipedia.org/wiki/… Euler's identity won't do?
    $endgroup$
    – mm-crj
    Dec 14 '18 at 17:54
















0












$begingroup$


I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?



For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?






derivatives complex-numbers polar-coordinates







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/… Euler's identity won't do?
    $endgroup$
    – mm-crj
    Dec 14 '18 at 17:54














0












0








0


0



$begingroup$


I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?



For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?






derivatives complex-numbers polar-coordinates







share|cite|improve this question









$endgroup$




I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?



For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?






derivatives complex-numbers polar-coordinates




derivatives complex-numbers polar-coordinates






share|cite|improve this question













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asked Dec 14 '18 at 17:48









Jonathan LowJonathan Low

635




635












  • $begingroup$
    en.wikipedia.org/wiki/… Euler's identity won't do?
    $endgroup$
    – mm-crj
    Dec 14 '18 at 17:54


















  • $begingroup$
    en.wikipedia.org/wiki/… Euler's identity won't do?
    $endgroup$
    – mm-crj
    Dec 14 '18 at 17:54
















$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54




$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54










2 Answers
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Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.






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    0












    $begingroup$

    Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:



    $ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,



    $ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      1












      $begingroup$

      Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.






          share|cite|improve this answer









          $endgroup$



          Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 19:03









          NarasimhamNarasimham

          20.8k52158




          20.8k52158























              0












              $begingroup$

              Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:



              $ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,



              $ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:



                $ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,



                $ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:



                  $ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,



                  $ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.






                  share|cite|improve this answer









                  $endgroup$



                  Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:



                  $ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,



                  $ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 18:25









                  KirtpoleKirtpole

                  1012




                  1012






























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