Question

Rezolvati in C ecuatia z^2-(2-i)z+3-i=0​

Answer 1

Răspuns:

Explicație pas cu pas:

Δ=(2-i)²-4·1·(3-i)=4-4i+i²-12+4i=-8+i²=-8+(-1)=-9=9·(-1)=9·i², deci √Δ=3i.

Atunci z1=(2-i-3i)/(2·1)=(2-4i)/2=(1-2i)·2/2=1-2i. z2=(2-i+3i)/2=(2+2i)/2=(1+i)·2/2=1+i.

Raspuns: S={1-2i; 1+i}

Answer 2

$z^2-(2-i)z+3-i = 0\\ \\ \Delta = (2-i)^2-4(3-i) = 4-4i-1-12+4i = -9\\ \\ z_{1,2} = \dfrac{2-i\pm \sqrt{9}i}{2} = \dfrac{2-i\pm 3i}{2}\\ \\ \Rightarrow z_1 = \dfrac{2-i+3i}{2} = \dfrac{2+2i}{2} = 1+i\\\\ \Rightarrow z_2 = \dfrac{2-i-3i}{2} = \dfrac{2-4i}{2} = 1-2i\\ \\\Rightarrow \boxed{S =\{1-2i; 1+i\}}$