Question

Fie z1 si z2 solutiile complexe ale ecuatiei 2z^2+z+50=0.Sa se calculeze |z1|+|z2|.

Answer 1

2z²  + z  + 50 = 0 
Δ = 1 - 400 = -399 = 399 · ( -1) = 399 · i² 
z₁ = ( -1  - √399 i ) / 4  =  - 1 /4   - √399i /4
         I z₁ I = radical mare = √( -1 /4)² +( -√399/4) = √1 /16 + 399/16 =√400 /√16
             = 20 /4 = 5 
identic   I z₂  I = 5

Answer 2

Ecuatia avand coeficientii reali si Δ<0 atunci ea are radacinile z1,z2 complexe, conjugate.$z_1=\overline{z_2}\Rightarrow |z_1|=|z_2|$
Din relatiile lui Viete avem ca :
[tex]z_1\cdot z_2=\frac{c}{a}=25\Rightarrow |z_1\cdot z_2|=25 \\ |z_1\cdot z_2|=|z_1||z_2|=|z_1||z_1|=|z_1|^2=25\Rightarrow |z_1|=|z_2|=5\Rightarrow\\ |z_1|+|z_2|=5+5=10[/tex]