Question

Rezolvati in multimea numerelor complexe ecuatia:
z^2=3-4i

Answer 1

$z^2=3-4i \\\\ \underline{z=ai+b} \\\\\\ (ai+b)^2= 3-4i \\\\ (ai)^2+2abi+b^2= 3-4i \\\\ -a^2+2abi+b^2= 3-4i \\\\ \left \{ {{-a^2+b^2=3} \ \ \ \ \ \ \ \atop {2ab=-4} \ \ \ \ \textless \ =\ \textgreater \ ab=-2} \right. \\\\\\ b^2=3+a^2 \\ b=\frac{-2}{a} \\\\\\ (\frac{-2}{a})^2=3+a^2 \\\\ \frac{4}{a^2}=3+a^2 \ \ \ |*a^2 \\\\ a^4+3a^2-4=0 \\\\ t\to a^2 \\\\ t^2+3t-4=0 \\\\\\ \Delta=9-4*(-4)=9+16\to 25 \\\\ t_1=\frac{-3+5}{2} \to 1 \\\\ t_2= \frac{-3-5}{2} \to -4 \\\\\\ a^2=1 \to a=1 \longrightarrow b=\frac{-2}{1} \to -2 \\\\ a^2=-4$

$a \to \sqrt{-4} = 2i \ \not \in \ R \\\\\\ \hbox{Ramane doar solutia pentru a=1 si b= -2 si obtinem numarul complex:} \\\\ \boxed{\boxed{z=i-2}}$