Question

sa se scrie ecuatia de gradul 3 cu coeficienti complecsi care are solutiile z1=1, z2=i, z3=1-i​

Answer 1

Răspuns:

nu ECUATIA, ci O ecuatie pt ca sunt o infinitate

Explicație pas cu pas:

conf. teoremei fundamentale  a algebrei, o scriem sub forma algebrica

a(z-1) (z-i)(z-1+i)=0 unde a∈ R*

dezvolti tu... ca sa obtii forma generala

are coeficienti complecsi, datorita factuluica rad.complexe nu sunt conjgate

Answer 2

 

$\displaystyle\bf\\Explicatii:\\O~ecuatie~la~care~cunosc~solutiile~se~scrie~cu~formula:\\\\(x-x_1)(x-x_2)(x-x_3)=0~~pentru~Real\\\\(z-z_1)(z-z_2)(z-z_3)=0~~pentru~complex.\\\\Rezolvare:\\\\(z-z_1)(z-z_2)(z-z_3)=0\\\\(z-1)(z-i)(z-(1-i))=0\\\\Desfacem~primele~2~paranteze.\\\\(z^2-iz-1z+1\times i)(z-(1-i))=0\\\\(z^2-(1+i)z+i)(z-(1-i))=0$

.

$\displaystyle\bf\\Desfacem~parantezele~ramase.\\\\z^3-(1+i)z^2+iz-(1-i)z^2+(1-i)(1+i)z -i(1-i)=0\\\\Facem~reducerile:\\\\z^3-[(1+i)+(1-i)]z^2+[i+(1-i)(1+i)]z -i(1-i)=0\\\\z^3-[1+i+1-i]z^2+[i+(1^2-i^2)]z+(-i+i^2)=0\\\\z^3-2z^2+[i+(1-(-1))]z+(-i-1)=0\\\\z^3-2z^2+[i+(1+1)]z-(1+i)=0\\\\\boxed{\bf z^3-2z^2+(2+i)z-(1+i)=0}$