Find real part of messy complex fraction?












-3












$begingroup$


I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?



$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$



where $R, Xin mathbb R$.






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, $Re(300)=300$
    $endgroup$
    – gammatester
    Nov 13 '18 at 10:39






  • 1




    $begingroup$
    Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
    $endgroup$
    – saulspatz
    Nov 13 '18 at 10:43
















-3












$begingroup$


I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?



$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$



where $R, Xin mathbb R$.






complex-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, $Re(300)=300$
    $endgroup$
    – gammatester
    Nov 13 '18 at 10:39






  • 1




    $begingroup$
    Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
    $endgroup$
    – saulspatz
    Nov 13 '18 at 10:43














-3












-3








-3





$begingroup$


I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?



$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$



where $R, Xin mathbb R$.






complex-numbers







share|cite|improve this question









$endgroup$




I have this messy complex fraction, where the LHS is real. But how can I find the real part of the RHS?



$$
frac{R277-X115+i(R115+X277)}{R+277+i(115+X)}=300,
$$



where $R, Xin mathbb R$.






complex-numbers




complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 13 '18 at 10:36









JDoeDoeJDoeDoe

7721614




7721614












  • $begingroup$
    Well, $Re(300)=300$
    $endgroup$
    – gammatester
    Nov 13 '18 at 10:39






  • 1




    $begingroup$
    Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
    $endgroup$
    – saulspatz
    Nov 13 '18 at 10:43


















  • $begingroup$
    Well, $Re(300)=300$
    $endgroup$
    – gammatester
    Nov 13 '18 at 10:39






  • 1




    $begingroup$
    Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
    $endgroup$
    – saulspatz
    Nov 13 '18 at 10:43
















$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39




$begingroup$
Well, $Re(300)=300$
$endgroup$
– gammatester
Nov 13 '18 at 10:39




1




1




$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43




$begingroup$
Does $R277$ mean the same thing as $277R?$ In any event, can't you multiply numerator and denominator by the complex conjugate of the denominator?
$endgroup$
– saulspatz
Nov 13 '18 at 10:43










2 Answers
2






active

oldest

votes


















0












$begingroup$

I believe that R=resistance and X=impedance.



So:



$R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$



begin{cases}
R277−X115 = 300 (R+277) \
R115+X277 = 300 (115+X) \
end{cases}



begin{cases}
R277−X115 = 300R+300*277 \
R115+X277 = 300*115+300X \
end{cases}



begin{cases}
R277-300R-X115 = 300*277 \
R115+X277-300X = 300*115 \
end{cases}



begin{cases}
-23R-115X = 300*277 = 83100 \
115R+577X = 300*115 = 34500 \
end{cases}



begin{cases}
R=-1128613 \
X= 225000 \
end{cases}



But R is negative! Can anybody verify ?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:



    $$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$



    Therefore, the real part of this fraction is



    $$ frac{ac+bd}{c^2+d^2}. $$



    In your case:




    • $a=R277-X115$

    • $b=R115+X277$

    • $c=R+277$

    • $d=X+115$


    Plugging these into the formula above, we get



    $$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



    Cancelling terms in the numerator gives



    $$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



    This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.




    In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.



    The resulting system is best solved by machine: here's the computation in WolframAlpha.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      I believe that R=resistance and X=impedance.



      So:



      $R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$



      begin{cases}
      R277−X115 = 300 (R+277) \
      R115+X277 = 300 (115+X) \
      end{cases}



      begin{cases}
      R277−X115 = 300R+300*277 \
      R115+X277 = 300*115+300X \
      end{cases}



      begin{cases}
      R277-300R-X115 = 300*277 \
      R115+X277-300X = 300*115 \
      end{cases}



      begin{cases}
      -23R-115X = 300*277 = 83100 \
      115R+577X = 300*115 = 34500 \
      end{cases}



      begin{cases}
      R=-1128613 \
      X= 225000 \
      end{cases}



      But R is negative! Can anybody verify ?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        I believe that R=resistance and X=impedance.



        So:



        $R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$



        begin{cases}
        R277−X115 = 300 (R+277) \
        R115+X277 = 300 (115+X) \
        end{cases}



        begin{cases}
        R277−X115 = 300R+300*277 \
        R115+X277 = 300*115+300X \
        end{cases}



        begin{cases}
        R277-300R-X115 = 300*277 \
        R115+X277-300X = 300*115 \
        end{cases}



        begin{cases}
        -23R-115X = 300*277 = 83100 \
        115R+577X = 300*115 = 34500 \
        end{cases}



        begin{cases}
        R=-1128613 \
        X= 225000 \
        end{cases}



        But R is negative! Can anybody verify ?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          I believe that R=resistance and X=impedance.



          So:



          $R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$



          begin{cases}
          R277−X115 = 300 (R+277) \
          R115+X277 = 300 (115+X) \
          end{cases}



          begin{cases}
          R277−X115 = 300R+300*277 \
          R115+X277 = 300*115+300X \
          end{cases}



          begin{cases}
          R277-300R-X115 = 300*277 \
          R115+X277-300X = 300*115 \
          end{cases}



          begin{cases}
          -23R-115X = 300*277 = 83100 \
          115R+577X = 300*115 = 34500 \
          end{cases}



          begin{cases}
          R=-1128613 \
          X= 225000 \
          end{cases}



          But R is negative! Can anybody verify ?






          share|cite|improve this answer









          $endgroup$



          I believe that R=resistance and X=impedance.



          So:



          $R277−X115+i(R115+X277) = 300 (R+277+i(115+X))$



          begin{cases}
          R277−X115 = 300 (R+277) \
          R115+X277 = 300 (115+X) \
          end{cases}



          begin{cases}
          R277−X115 = 300R+300*277 \
          R115+X277 = 300*115+300X \
          end{cases}



          begin{cases}
          R277-300R-X115 = 300*277 \
          R115+X277-300X = 300*115 \
          end{cases}



          begin{cases}
          -23R-115X = 300*277 = 83100 \
          115R+577X = 300*115 = 34500 \
          end{cases}



          begin{cases}
          R=-1128613 \
          X= 225000 \
          end{cases}



          But R is negative! Can anybody verify ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 21:06









          cgiovanardicgiovanardi

          751411




          751411























              0












              $begingroup$

              I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:



              $$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$



              Therefore, the real part of this fraction is



              $$ frac{ac+bd}{c^2+d^2}. $$



              In your case:




              • $a=R277-X115$

              • $b=R115+X277$

              • $c=R+277$

              • $d=X+115$


              Plugging these into the formula above, we get



              $$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



              Cancelling terms in the numerator gives



              $$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



              This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.




              In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.



              The resulting system is best solved by machine: here's the computation in WolframAlpha.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:



                $$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$



                Therefore, the real part of this fraction is



                $$ frac{ac+bd}{c^2+d^2}. $$



                In your case:




                • $a=R277-X115$

                • $b=R115+X277$

                • $c=R+277$

                • $d=X+115$


                Plugging these into the formula above, we get



                $$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                Cancelling terms in the numerator gives



                $$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.




                In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.



                The resulting system is best solved by machine: here's the computation in WolframAlpha.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:



                  $$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$



                  Therefore, the real part of this fraction is



                  $$ frac{ac+bd}{c^2+d^2}. $$



                  In your case:




                  • $a=R277-X115$

                  • $b=R115+X277$

                  • $c=R+277$

                  • $d=X+115$


                  Plugging these into the formula above, we get



                  $$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                  Cancelling terms in the numerator gives



                  $$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                  This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.




                  In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.



                  The resulting system is best solved by machine: here's the computation in WolframAlpha.






                  share|cite|improve this answer









                  $endgroup$



                  I'm assuming that you flipped right and left (since the real part of $300$ is clearly $300$). In general, to put a complex fraction in standard $x+yi$ form, you can multiply both the numerator and denominator by the conjugate of the denominator (as saulspatz said in their comment) In symbols:



                  $$ frac{a+bi}{c+di} ~~=~~ frac{a+bi}{c+di} cdot frac{c-di}{c-di} ~~=~~ frac{ac+bd + (bc-ad)i}{c^2+d^2}.$$



                  Therefore, the real part of this fraction is



                  $$ frac{ac+bd}{c^2+d^2}. $$



                  In your case:




                  • $a=R277-X115$

                  • $b=R115+X277$

                  • $c=R+277$

                  • $d=X+115$


                  Plugging these into the formula above, we get



                  $$frac{[R^2277-RX115+R277^2-X(277cdot115)]+[RX115+X^2277+R115^2+X(277cdot115)]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                  Cancelling terms in the numerator gives



                  $$=frac{[R^2277+R277^2]+[X^2277+R115^2]}{R^2+2R277+277^2+X^2+2X115+115^2}$$



                  This is still a mess, but it doesn't seem like you can do much better except to write out some of the numbers e.g. $115^2=13225$, and so on.




                  In the other answer, cgiovanardi assumed that your ultimate goal was to solve this equation, in which case their idea is a lot better. Rather than getting an explicit form for the real and imaginary part of the LHS as a fraction, you can multiply the denominator through to both sides first.



                  The resulting system is best solved by machine: here's the computation in WolframAlpha.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 4:57









                  aleph_twoaleph_two

                  24912




                  24912






























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