Se considera trei numere complexe \( |z_{1}|=|z_{2}|=|z_{3}| \)
a) Dem. ca \( (\exists) \) a, b nr. complexe a.i. \( z_{2}=az_{1},\ z_{3}=bz_{1}, |a|=|b|=1 \).
b)Rezolvati ecuatia \( a^2+b^2-ab-a-b+1=0 \) in raport cu una din necunoscutele a sau b.
Trei numere complexe
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mihai miculita
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a) \( {\mbox Daca }
\left\{\begin{array} z_1=|z_1|(\cos{t_1}+i\sin{t_1})\\
z_2=|z_2|(\cos{t_2}+i\sin{t_2})\\
z_3=|z_3|(\cos{t_3}+i\sin{t_3}) \end{\array}
\Rightarrow \left\{\begin{\array}
\frac{z_2}{z_1}=\frac{|z_2|}{|z_1|}[\cos(t_2-t_1)+i\sin(t_2-t_1)]=\cos(t_2-t_1)+i\sin(t_2-t_1)=a\in\mathbb{C}\\
\frac{z_3}{z_1}=\frac{|z_3|}{|z_1|}[\cos(t_3-t_1)+i\sin(t_3-t_1)]=\cos(t_3-t_1)+i\sin(t_3-t_1)=b\in\mathbb{C}. \)
b) \( b^2-(a+1)b+(a^2-a+1)=0\Rightarrow \triangle =(a+1)^2-4(a^2-a+1)=-3a^2+6a-3=-3(a-1)^2\Rightarrow \) \( b_{1;2}=\frac{a+1{\pm}i(a-1)\sqrt{3}}{2}. \)
\left\{\begin{array} z_1=|z_1|(\cos{t_1}+i\sin{t_1})\\
z_2=|z_2|(\cos{t_2}+i\sin{t_2})\\
z_3=|z_3|(\cos{t_3}+i\sin{t_3}) \end{\array}
\Rightarrow \left\{\begin{\array}
\frac{z_2}{z_1}=\frac{|z_2|}{|z_1|}[\cos(t_2-t_1)+i\sin(t_2-t_1)]=\cos(t_2-t_1)+i\sin(t_2-t_1)=a\in\mathbb{C}\\
\frac{z_3}{z_1}=\frac{|z_3|}{|z_1|}[\cos(t_3-t_1)+i\sin(t_3-t_1)]=\cos(t_3-t_1)+i\sin(t_3-t_1)=b\in\mathbb{C}. \)
b) \( b^2-(a+1)b+(a^2-a+1)=0\Rightarrow \triangle =(a+1)^2-4(a^2-a+1)=-3a^2+6a-3=-3(a-1)^2\Rightarrow \) \( b_{1;2}=\frac{a+1{\pm}i(a-1)\sqrt{3}}{2}. \)