Question

Să se determine numerele complexe z pentru care:
c) (1-i) z² + (1+i)z+2=2i; ​

Answer 1

$\it (1-i)z^2+(1+i)z+2=2i \Rightarrow (1-i)z^2+(1+i)z+2-2i=0 \Rightarrow \\ \\ \\ \Rightarrow (1-i)z^2+(1+i)z+2(1-i)=0\bigg|_{\cdot(1+i)} \Rightarrow 2z^2+2iz+2\cdot2=0\bigg|_{:2} \Rightarrow \\ \\ \\ \Rightarrow z^2+iz+2=0\\ \\ \\ \Delta = i^2-4\cdot2=-1-8=-9\\ \\ \\ z_{1,2}=\dfrac{-i\pm\sqrt{-9}}{2}=\dfrac{-i\pm3i}{2} \Rightarrow\begin{cases} \it z_1=\dfrac{-i-3i}{2}=\dfrac{-4i}{2}=-2i\\ \\ \\ \it z_2=\dfrac{-i+3i}{2}=\dfrac{2i}{2}=i\end{cases}$

Answer 2

Răspuns:

-2i si i

Explicație pas cu pas:

(1-i)z²+(1+i)z+2(1-i)=0

Impartim cu (1-i)

z²+(1+i)/(1-i)z+2=0

Prelucram (1+i)/(1-i)=(1+i)²/(1-i²)=(1+2i-1)/(1+1)=2i/2=i

Deci ecuatia devine:

z²+iz+2=0

z12=(-i +- rad(i²-8))/2=(-1 +- 3i)/2

Sau z1=-2i si z2=i