[tex]\text{Fie }z=a+bi,~~~a,b\in \mathbb{R}\\z^2=10+6i\\(a+bi)^2=10+6i\\a^2+2abi-b^2=10+6i\\\text{Identificand coeficientii obtinem sistemul:}\\\left \{ {{a^2-b^2=10} \atop {2ab=6}} \right.\Leftrightarrow \left \{ {{a^2-b^2=10} \atop {ab=3}} \right.[/tex]
Din a doua relatie obtinem:
[tex]a=\dfrac{3}{b}[/tex]
Inlocuim in prima relatie:
[tex]\dfrac{9}{b^2}-b^2=10\\9-b^4=10b^2\\b^4+10b^2-9=0\\b^2\stackrel{not}{=} t,t>0\\t^2+10t-9=0\\\Delta= 100+36=136\Rightarrow \sqrt{\Delta} =2\sqrt{34}\\t_1=\dfrac{-10+2\sqrt{34}}{2} = -5+\sqrt{34}\\t_2=\dfrac{-10-2\sqrt{34}}{2}= -5-\sqrt {34} <0,\text{deci nu convine}\\b^2=\sqrt{34}-5 \Rightarrow b=\pm \sqrt{\sqrt{34}-5}\\\text{De aici il scoti pe a.}[/tex]
