Question

z²+2z_+1=0

rezolvați ecuația

2z_ este 2zconjugat ​

Answer 1

Răspuns:

Fie z = $a+bi$

$=>(a+bi)^2+2(a-bi)+1=0$

$<=>a^2-b^2+2abi+2a-2bi+1=0=>a^2-b^2+2a+1=0$

si $2ab-2b=0=>2b(a-1)=0$

$Caz 1: b=0=>a^2-0+2a+1=0=>a=-1 =>z1=-1$

$Caz 2:a=1=>1-b^2+2+1=0>b^2=-4=>b=$±$2i=> z2=1+2i$

$z3=1-2i$