Prove $frac{1+iz}{1-iz}=frac{a+ib}{1+c}$ if $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$

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If $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$ then prove that $dfrac{1+iz}{1-iz}=dfrac{a+ib}{1+c}$




My Attempt
$$
z=frac{b}{1+a}+ifrac{c}{1+a}implies iz=frac{-c}{1+a}+ifrac{b}{1+a}\
frac{1+iz}{1-iz}=frac{frac{1+a-c}{1+a}+ifrac{b}{1+a}}{frac{1+a+c}{1+a}-ifrac{b}{1+a}}=frac{1+a-c+ib}{1+a+c-ib}.frac{1+a+c+ib}{1+a+c+ib}\
=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}=frac{1+a^2-b^2+2a+2iab+2ib-c^2}{1+a^2+c^2+2a+2c+2ac+b^2}
$$
I don't think its going anywhere with my attempt, how can I solve it easily as it was asked as a multiple choice question ?






complex-numbers







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  • $begingroup$
    How is a prove that question multiple choice?
    $endgroup$
    – Peter Foreman
    Jan 22 at 8:13










  • $begingroup$
    @PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
    $endgroup$
    – ss1729
    Jan 22 at 8:14








  • 1




    $begingroup$
    books.google.co.in/…
    $endgroup$
    – lab bhattacharjee
    Jan 22 at 9:11
















1












$begingroup$



If $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$ then prove that $dfrac{1+iz}{1-iz}=dfrac{a+ib}{1+c}$




My Attempt
$$
z=frac{b}{1+a}+ifrac{c}{1+a}implies iz=frac{-c}{1+a}+ifrac{b}{1+a}\
frac{1+iz}{1-iz}=frac{frac{1+a-c}{1+a}+ifrac{b}{1+a}}{frac{1+a+c}{1+a}-ifrac{b}{1+a}}=frac{1+a-c+ib}{1+a+c-ib}.frac{1+a+c+ib}{1+a+c+ib}\
=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}=frac{1+a^2-b^2+2a+2iab+2ib-c^2}{1+a^2+c^2+2a+2c+2ac+b^2}
$$
I don't think its going anywhere with my attempt, how can I solve it easily as it was asked as a multiple choice question ?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    How is a prove that question multiple choice?
    $endgroup$
    – Peter Foreman
    Jan 22 at 8:13










  • $begingroup$
    @PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
    $endgroup$
    – ss1729
    Jan 22 at 8:14








  • 1




    $begingroup$
    books.google.co.in/…
    $endgroup$
    – lab bhattacharjee
    Jan 22 at 9:11














1












1








1


2



$begingroup$



If $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$ then prove that $dfrac{1+iz}{1-iz}=dfrac{a+ib}{1+c}$




My Attempt
$$
z=frac{b}{1+a}+ifrac{c}{1+a}implies iz=frac{-c}{1+a}+ifrac{b}{1+a}\
frac{1+iz}{1-iz}=frac{frac{1+a-c}{1+a}+ifrac{b}{1+a}}{frac{1+a+c}{1+a}-ifrac{b}{1+a}}=frac{1+a-c+ib}{1+a+c-ib}.frac{1+a+c+ib}{1+a+c+ib}\
=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}=frac{1+a^2-b^2+2a+2iab+2ib-c^2}{1+a^2+c^2+2a+2c+2ac+b^2}
$$
I don't think its going anywhere with my attempt, how can I solve it easily as it was asked as a multiple choice question ?






complex-numbers







share|cite|improve this question











$endgroup$





If $b+ic=(1+a)z$ and $a^2+b^2+c^2=1$ then prove that $dfrac{1+iz}{1-iz}=dfrac{a+ib}{1+c}$




My Attempt
$$
z=frac{b}{1+a}+ifrac{c}{1+a}implies iz=frac{-c}{1+a}+ifrac{b}{1+a}\
frac{1+iz}{1-iz}=frac{frac{1+a-c}{1+a}+ifrac{b}{1+a}}{frac{1+a+c}{1+a}-ifrac{b}{1+a}}=frac{1+a-c+ib}{1+a+c-ib}.frac{1+a+c+ib}{1+a+c+ib}\
=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}=frac{1+a^2-b^2+2a+2iab+2ib-c^2}{1+a^2+c^2+2a+2c+2ac+b^2}
$$
I don't think its going anywhere with my attempt, how can I solve it easily as it was asked as a multiple choice question ?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 9:44









greedoid

44.6k1156111




44.6k1156111










asked Jan 22 at 8:09









ss1729ss1729

1,97511023




1,97511023












  • $begingroup$
    How is a prove that question multiple choice?
    $endgroup$
    – Peter Foreman
    Jan 22 at 8:13










  • $begingroup$
    @PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
    $endgroup$
    – ss1729
    Jan 22 at 8:14








  • 1




    $begingroup$
    books.google.co.in/…
    $endgroup$
    – lab bhattacharjee
    Jan 22 at 9:11


















  • $begingroup$
    How is a prove that question multiple choice?
    $endgroup$
    – Peter Foreman
    Jan 22 at 8:13










  • $begingroup$
    @PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
    $endgroup$
    – ss1729
    Jan 22 at 8:14








  • 1




    $begingroup$
    books.google.co.in/…
    $endgroup$
    – lab bhattacharjee
    Jan 22 at 9:11
















$begingroup$
How is a prove that question multiple choice?
$endgroup$
– Peter Foreman
Jan 22 at 8:13




$begingroup$
How is a prove that question multiple choice?
$endgroup$
– Peter Foreman
Jan 22 at 8:13












$begingroup$
@PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
$endgroup$
– ss1729
Jan 22 at 8:14






$begingroup$
@PeterForeman it was to evaluate $frac{1+iz}{1-iz}$, and the solution given in my reference was $frac{a+ib}{1+c}$
$endgroup$
– ss1729
Jan 22 at 8:14






1




1




$begingroup$
books.google.co.in/…
$endgroup$
– lab bhattacharjee
Jan 22 at 9:11




$begingroup$
books.google.co.in/…
$endgroup$
– lab bhattacharjee
Jan 22 at 9:11










1 Answer
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$begingroup$

begin{align}
frac{1+iz}{1-iz}&=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\
&=frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\
&=frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\
&=frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\
&=frac{a^2+a+iab+ib}{1+a+c+ac}\
&=frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\
&=frac{a+ib}{1+c}\
end{align}



Can you find out your mistake now?



If you are interested, see spherical representation of complex numbers.






share|cite|improve this answer











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    $begingroup$

    begin{align}
    frac{1+iz}{1-iz}&=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\
    &=frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\
    &=frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\
    &=frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\
    &=frac{a^2+a+iab+ib}{1+a+c+ac}\
    &=frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\
    &=frac{a+ib}{1+c}\
    end{align}



    Can you find out your mistake now?



    If you are interested, see spherical representation of complex numbers.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      begin{align}
      frac{1+iz}{1-iz}&=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\
      &=frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\
      &=frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\
      &=frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\
      &=frac{a^2+a+iab+ib}{1+a+c+ac}\
      &=frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\
      &=frac{a+ib}{1+c}\
      end{align}



      Can you find out your mistake now?



      If you are interested, see spherical representation of complex numbers.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        begin{align}
        frac{1+iz}{1-iz}&=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\
        &=frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\
        &=frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\
        &=frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\
        &=frac{a^2+a+iab+ib}{1+a+c+ac}\
        &=frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\
        &=frac{a+ib}{1+c}\
        end{align}



        Can you find out your mistake now?



        If you are interested, see spherical representation of complex numbers.






        share|cite|improve this answer











        $endgroup$



        begin{align}
        frac{1+iz}{1-iz}&=frac{(1+a+ib)^2-c^2}{(1+a+c)^2+b^2}\
        &=frac{1+a^2-b^2-c^2+2a+2iab+2ib}{1+a^2+b^2+c^2+2a+2c+2ac}\
        &=frac{1+a^2-(1-a^2)+2a+2iab+2ib}{1+1+2a+2c+2ac}\
        &=frac{2a^2+2a+2iab+2ib}{2+2a+2c+2ac}\
        &=frac{a^2+a+iab+ib}{1+a+c+ac}\
        &=frac{a(a+1)+ib(a+1)}{(1+a)+c(1+a)}\
        &=frac{a+ib}{1+c}\
        end{align}



        Can you find out your mistake now?



        If you are interested, see spherical representation of complex numbers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 9:20









        Martin R

        29.4k33558




        29.4k33558










        answered Jan 22 at 9:08









        Thomas ShelbyThomas Shelby

        3,6992525




        3,6992525






























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