Question

Ma poate ajuta cineva cu aceasta problema?

Answer 1

z⁶ - 9z³ + 8 = z⁶ - z³ - 8z³ + 8 = z³(z³-1) - 8(z³ - 1)=(z³ - 1)(z³ - 8).

z³ - 1 = 0, deci z³ = 1

$z^3=1\Rightarrow z=cos\dfrac{2k\pi}{3}+i \cdot sin\dfrac{2k\pi}{3},\;unde\;k=0,\;1,\;2,\;deci\\\\z_1=1,\;z_2=cos\dfrac{2\pi}3+i\cdot sin\dfrac{2\pi}3=-\dfrac{1}2+i\cdot\dfrac{\sqrt3}2,\;z_3=-\dfrac{1}2-i\cdot\dfrac{\sqrt3}2;\\\\z^3=8\Rightarrow z=2\left(cos\dfrac{2k\pi}{3}+i \cdot sin\dfrac{2k\pi}{3}\right),\;unde\;k=0,\;1,\;2,\;deci\\\\z_4=2,\;z_5=2\left(cos\dfrac{2\pi}3+i\cdot sin\dfrac{2\pi}3\right)=2\left(-\dfrac{1}2+i\cdot\dfrac{\sqrt3}2\right)=-1+i\cdot\sqrt3,\\\\z_6=-1-i\cdot\sqrt3.$

Răspunsul corect este deci F.

Green eyes.