Find the geometric locus $z in mathbb C$ so that $frac{z+2}{z(z+1)}in mathbb R$












4












$begingroup$



Find the geometric locus of the set of $z in mathbb C$ so that
$$frac{z+2}{z(z+1)}in mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)




My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).



Hints and solutions are welcomed.






algebra-precalculus complex-numbers contest-math







share|cite|improve this question









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    4












    $begingroup$



    Find the geometric locus of the set of $z in mathbb C$ so that
    $$frac{z+2}{z(z+1)}in mathbb R$$
    Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)




    My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).



    Hints and solutions are welcomed.






    algebra-precalculus complex-numbers contest-math







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      0



      $begingroup$



      Find the geometric locus of the set of $z in mathbb C$ so that
      $$frac{z+2}{z(z+1)}in mathbb R$$
      Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)




      My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).



      Hints and solutions are welcomed.






      algebra-precalculus complex-numbers contest-math







      share|cite|improve this question









      $endgroup$





      Find the geometric locus of the set of $z in mathbb C$ so that
      $$frac{z+2}{z(z+1)}in mathbb R$$
      Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)




      My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$ but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).



      Hints and solutions are welcomed.






      algebra-precalculus complex-numbers contest-math




      algebra-precalculus complex-numbers contest-math






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 10:48









      bluemasterbluemaster

      1,663520




      1,663520






















          4 Answers
          4






          active

          oldest

          votes


















          7












          $begingroup$

          A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
          $$
          frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
          $$
          Cross multiplication gives
          $$
          zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
          $$
          and we can transfer everything to the right-hand side:
          $$
          zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
          $$
          This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
          $$
          zbar{z}+2(z+bar{z})+2=0 tag{*}
          $$
          Now we can substitute $z=x+yi$:
          $$
          x^2+y^2+4x+2=0
          $$
          that is,
          $$
          (x+2)^2+y^2=2
          $$
          which is a circle.



          You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
          $$
          |z+2|^2=2
          $$
          which is clearly a circle with center at $2$ and radius $sqrt{2}$.






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            HINT: Multiply the fraction by
            $$
            frac{bar z(bar z-1)}{bar z(bar z-1)};.
            $$
            In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.



              $$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$



              Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$



              and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$



              Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$



              $iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$






              share|cite|improve this answer









              $endgroup$





















                2












                $begingroup$

                Using



                $$
                z = sqrt{x^2+y^2}e^{arctanfrac yx}
                $$



                we have



                $$
                frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
                $$



                and we seek for



                $$
                arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
                $$



                then



                $$
                tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
                $$



                or



                $$
                y(y^2+x^2+4x+2) = 0
                $$



                NOTE



                $$
                tan(a-b) = frac{tan a-tan b}{1+tan atan b}
                $$



                so



                $$
                tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
                $$



                etc.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  How did you get that last step?
                  $endgroup$
                  – I like Serena
                  Dec 16 '18 at 14:37






                • 2




                  $begingroup$
                  @IlikeSerena Please. See attached note.
                  $endgroup$
                  – Cesareo
                  Dec 16 '18 at 16:02










                • $begingroup$
                  @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                  $endgroup$
                  – Cesareo
                  Dec 19 '18 at 11:57











                Your Answer





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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                7












                $begingroup$

                A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
                $$
                frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
                $$
                Cross multiplication gives
                $$
                zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
                $$
                and we can transfer everything to the right-hand side:
                $$
                zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
                $$
                This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
                $$
                zbar{z}+2(z+bar{z})+2=0 tag{*}
                $$
                Now we can substitute $z=x+yi$:
                $$
                x^2+y^2+4x+2=0
                $$
                that is,
                $$
                (x+2)^2+y^2=2
                $$
                which is a circle.



                You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
                $$
                |z+2|^2=2
                $$
                which is clearly a circle with center at $2$ and radius $sqrt{2}$.






                share|cite|improve this answer











                $endgroup$


















                  7












                  $begingroup$

                  A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
                  $$
                  frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
                  $$
                  Cross multiplication gives
                  $$
                  zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
                  $$
                  and we can transfer everything to the right-hand side:
                  $$
                  zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
                  $$
                  This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
                  $$
                  zbar{z}+2(z+bar{z})+2=0 tag{*}
                  $$
                  Now we can substitute $z=x+yi$:
                  $$
                  x^2+y^2+4x+2=0
                  $$
                  that is,
                  $$
                  (x+2)^2+y^2=2
                  $$
                  which is a circle.



                  You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
                  $$
                  |z+2|^2=2
                  $$
                  which is clearly a circle with center at $2$ and radius $sqrt{2}$.






                  share|cite|improve this answer











                  $endgroup$
















                    7












                    7








                    7





                    $begingroup$

                    A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
                    $$
                    frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
                    $$
                    Cross multiplication gives
                    $$
                    zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
                    $$
                    and we can transfer everything to the right-hand side:
                    $$
                    zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
                    $$
                    This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
                    $$
                    zbar{z}+2(z+bar{z})+2=0 tag{*}
                    $$
                    Now we can substitute $z=x+yi$:
                    $$
                    x^2+y^2+4x+2=0
                    $$
                    that is,
                    $$
                    (x+2)^2+y^2=2
                    $$
                    which is a circle.



                    You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
                    $$
                    |z+2|^2=2
                    $$
                    which is clearly a circle with center at $2$ and radius $sqrt{2}$.






                    share|cite|improve this answer











                    $endgroup$



                    A complex number $w$ is real if and only if it equals its conjugate $bar{w}$. So you need to solve
                    $$
                    frac{z+2}{z(z+1)}=frac{bar{z}+2}{bar{z}(bar{z}+1)}
                    $$
                    Cross multiplication gives
                    $$
                    zbar{z}^2+zbar{z}+2bar{z}^2+2bar{z}=z^2bar{z}+zbar{z}+2z^2+2z
                    $$
                    and we can transfer everything to the right-hand side:
                    $$
                    zbar{z}(z-bar{z})+2(z^2-bar{z}^2)+2(z-bar{z})=0
                    $$
                    This has the entire real line ($z=bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
                    $$
                    zbar{z}+2(z+bar{z})+2=0 tag{*}
                    $$
                    Now we can substitute $z=x+yi$:
                    $$
                    x^2+y^2+4x+2=0
                    $$
                    that is,
                    $$
                    (x+2)^2+y^2=2
                    $$
                    which is a circle.



                    You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $zbar{z}+2z+2bar{z}+4=2$, so $(z+2)(bar{z}+2)=2$ and finally
                    $$
                    |z+2|^2=2
                    $$
                    which is clearly a circle with center at $2$ and radius $sqrt{2}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 16 '18 at 12:17

























                    answered Dec 16 '18 at 10:59









                    egregegreg

                    184k1486205




                    184k1486205























                        4












                        $begingroup$

                        HINT: Multiply the fraction by
                        $$
                        frac{bar z(bar z-1)}{bar z(bar z-1)};.
                        $$
                        In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          HINT: Multiply the fraction by
                          $$
                          frac{bar z(bar z-1)}{bar z(bar z-1)};.
                          $$
                          In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            HINT: Multiply the fraction by
                            $$
                            frac{bar z(bar z-1)}{bar z(bar z-1)};.
                            $$
                            In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.






                            share|cite|improve this answer









                            $endgroup$



                            HINT: Multiply the fraction by
                            $$
                            frac{bar z(bar z-1)}{bar z(bar z-1)};.
                            $$
                            In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 16 '18 at 10:53









                            JoeJoe

                            7,21421229




                            7,21421229























                                2












                                $begingroup$

                                A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.



                                $$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$



                                Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$



                                and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$



                                Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$



                                $iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$






                                share|cite|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.



                                  $$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$



                                  Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$



                                  and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$



                                  Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$



                                  $iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.



                                    $$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$



                                    Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$



                                    and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$



                                    Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$



                                    $iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$






                                    share|cite|improve this answer









                                    $endgroup$



                                    A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation.



                                    $$dfrac{z+2}{z(z+1)}=dfrac1{z+1}+dfrac{2(z+1-z)}{z(z+1)}=dfrac2z-dfrac1{z+1}$$



                                    Now the imaginary part of $dfrac2z$ is $-dfrac{2y}{x^2+y^2}$



                                    and that of $dfrac1{z+1}$ is $-dfrac y{(x+1)^2+y^2}$



                                    Finally, $dfrac{z+2}{z(z+1)}$ will be real if $-dfrac{2y}{x^2+y^2}=-dfrac y{(x+1)^2+y^2}$



                                    $iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y{(x+2)^2+y^2-2}$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 16 '18 at 12:32









                                    lab bhattacharjeelab bhattacharjee

                                    226k15158275




                                    226k15158275























                                        2












                                        $begingroup$

                                        Using



                                        $$
                                        z = sqrt{x^2+y^2}e^{arctanfrac yx}
                                        $$



                                        we have



                                        $$
                                        frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
                                        $$



                                        and we seek for



                                        $$
                                        arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
                                        $$



                                        then



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
                                        $$



                                        or



                                        $$
                                        y(y^2+x^2+4x+2) = 0
                                        $$



                                        NOTE



                                        $$
                                        tan(a-b) = frac{tan a-tan b}{1+tan atan b}
                                        $$



                                        so



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
                                        $$



                                        etc.






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          How did you get that last step?
                                          $endgroup$
                                          – I like Serena
                                          Dec 16 '18 at 14:37






                                        • 2




                                          $begingroup$
                                          @IlikeSerena Please. See attached note.
                                          $endgroup$
                                          – Cesareo
                                          Dec 16 '18 at 16:02










                                        • $begingroup$
                                          @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                          $endgroup$
                                          – Cesareo
                                          Dec 19 '18 at 11:57
















                                        2












                                        $begingroup$

                                        Using



                                        $$
                                        z = sqrt{x^2+y^2}e^{arctanfrac yx}
                                        $$



                                        we have



                                        $$
                                        frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
                                        $$



                                        and we seek for



                                        $$
                                        arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
                                        $$



                                        then



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
                                        $$



                                        or



                                        $$
                                        y(y^2+x^2+4x+2) = 0
                                        $$



                                        NOTE



                                        $$
                                        tan(a-b) = frac{tan a-tan b}{1+tan atan b}
                                        $$



                                        so



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
                                        $$



                                        etc.






                                        share|cite|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          How did you get that last step?
                                          $endgroup$
                                          – I like Serena
                                          Dec 16 '18 at 14:37






                                        • 2




                                          $begingroup$
                                          @IlikeSerena Please. See attached note.
                                          $endgroup$
                                          – Cesareo
                                          Dec 16 '18 at 16:02










                                        • $begingroup$
                                          @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                          $endgroup$
                                          – Cesareo
                                          Dec 19 '18 at 11:57














                                        2












                                        2








                                        2





                                        $begingroup$

                                        Using



                                        $$
                                        z = sqrt{x^2+y^2}e^{arctanfrac yx}
                                        $$



                                        we have



                                        $$
                                        frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
                                        $$



                                        and we seek for



                                        $$
                                        arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
                                        $$



                                        then



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
                                        $$



                                        or



                                        $$
                                        y(y^2+x^2+4x+2) = 0
                                        $$



                                        NOTE



                                        $$
                                        tan(a-b) = frac{tan a-tan b}{1+tan atan b}
                                        $$



                                        so



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
                                        $$



                                        etc.






                                        share|cite|improve this answer











                                        $endgroup$



                                        Using



                                        $$
                                        z = sqrt{x^2+y^2}e^{arctanfrac yx}
                                        $$



                                        we have



                                        $$
                                        frac{z+2}{z(z+1)} = frac{rho_1 e^{iphi_1}}{rho_2 e^{iphi_2}rho_3 e^{iphi_3}}
                                        $$



                                        and we seek for



                                        $$
                                        arctanfrac{y}{x+2}-arctanfrac yx -arctanfrac{y}{x+1} = 0
                                        $$



                                        then



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac yx
                                        $$



                                        or



                                        $$
                                        y(y^2+x^2+4x+2) = 0
                                        $$



                                        NOTE



                                        $$
                                        tan(a-b) = frac{tan a-tan b}{1+tan atan b}
                                        $$



                                        so



                                        $$
                                        tanleft(arctanfrac{y}{x+2} -arctanfrac{y}{x+1}right) = frac{y}{y^2+x^2+3x+2}
                                        $$



                                        etc.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 19 '18 at 11:53

























                                        answered Dec 16 '18 at 14:02









                                        CesareoCesareo

                                        9,3063517




                                        9,3063517












                                        • $begingroup$
                                          How did you get that last step?
                                          $endgroup$
                                          – I like Serena
                                          Dec 16 '18 at 14:37






                                        • 2




                                          $begingroup$
                                          @IlikeSerena Please. See attached note.
                                          $endgroup$
                                          – Cesareo
                                          Dec 16 '18 at 16:02










                                        • $begingroup$
                                          @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                          $endgroup$
                                          – Cesareo
                                          Dec 19 '18 at 11:57


















                                        • $begingroup$
                                          How did you get that last step?
                                          $endgroup$
                                          – I like Serena
                                          Dec 16 '18 at 14:37






                                        • 2




                                          $begingroup$
                                          @IlikeSerena Please. See attached note.
                                          $endgroup$
                                          – Cesareo
                                          Dec 16 '18 at 16:02










                                        • $begingroup$
                                          @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                          $endgroup$
                                          – Cesareo
                                          Dec 19 '18 at 11:57
















                                        $begingroup$
                                        How did you get that last step?
                                        $endgroup$
                                        – I like Serena
                                        Dec 16 '18 at 14:37




                                        $begingroup$
                                        How did you get that last step?
                                        $endgroup$
                                        – I like Serena
                                        Dec 16 '18 at 14:37




                                        2




                                        2




                                        $begingroup$
                                        @IlikeSerena Please. See attached note.
                                        $endgroup$
                                        – Cesareo
                                        Dec 16 '18 at 16:02




                                        $begingroup$
                                        @IlikeSerena Please. See attached note.
                                        $endgroup$
                                        – Cesareo
                                        Dec 16 '18 at 16:02












                                        $begingroup$
                                        @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                        $endgroup$
                                        – Cesareo
                                        Dec 19 '18 at 11:57




                                        $begingroup$
                                        @bluemaster Thanks for the hint. I have a real problem with addition (and subtraction)
                                        $endgroup$
                                        – Cesareo
                                        Dec 19 '18 at 11:57


















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