Question

Fiind date numerele complexe dinstincte a,b,c astfel incat |a| = |b| = |c| si |b+c-a|=|a|, sa se arate ca b+c=0.

Answer 1

Salut,



[tex]Solutie.~Fie~a=r(cosx+isinx),b=r(cosy+isiny)~si~ c=r(cosz+isinz). Conditia~ /b+c-a/ = /a/ se ~scrie~succesiv~:~/r[(cosy+cosz-cosx)+ +i(siny+sinz-sinx)/]=r \ \textless \ =\ \textgreater \ (cosy+cosz-cos x)^2+(siny+sinz-sinx)^2 =1 \ \textless \ =\ \textgreater \ (cosy+cosz)^2+cos^2x +(siny+sinz)^2+sin^2x-2cosx(cosy+ +cosz)-2sinx(siny+sinz)=1 \ \textless \ =\ \textgreater \ (cosy+cosz)^2+(siny+sinz)^2=2cos ~ (x-y) + 2cos(x-z) \ \textless \ =\ \textgreater \ 2 + 2cos(y-z)=2cos(x-y) + 2cos(x-z) \ \textless \ =\ \textgreater \ \ \textless \ =\ \textgreater \ 1 + cos(y-z)=cos(x-y)+cos(x-z) \ \textless \ =\ \textgreater \ \ \textless \ =\ \textgreater \ 2cos^2 ~\dfrac{y-z}{2} =2cos\dfrac{y-z}{2}cos \dfrac{2x-y-z}{2} \ \textless \ =\ \textgreater \ 

\ \textless \ =\ \textgreater \ 2cos~\dfrac{y-z}{2}(cos\dfrac{y-z}{2}-cos\dfrac{2x-y-z}{2})=0 \ \textless \ =\ \textgreater \ \ \textless \ =\ \textgreater \ 4cos\dfrac{y-z}{2} sin \dfrac{x-z}{2} sin \dfrac{x-y}{2}=0, (1),~ Din~(1)~rezulta~cos \dfrac{y-z}{2}=0, ~(2), ~sau~sin\dfrac{x-z}{2}= =0, ~(3),~sau~sin\dfrac{x-y}{2}=0, ~(4). Dar~b+c=r(cosy+i~siny)+r(cosz+i~sinz)~sau b+c=r(2cos\dfrac{y+z}{2}cos\dfrac{y-z}{2}+2i ~sin\dfrac{y+z}{2} cos\dfrac{y-z}{2})~sau~b+c=2r~cos\dfrac{y-z}{2}(cos\dfrac{y+z}{2}+ +i~sin\dfrac{y+z}{2}),~(5).[/tex]

Tinand seama de egalitatea (2) din (5) rezulta ca b + c = 0.

Egalitatile (3) si (4) nu pot avea loc deoarece prin ipoteza problemei, numerele a, b, c sunt distincte.