Matematyka.pl

zad. 1 \(\displaystyle{ x^{2}= (1+8i)x + 16 - 4i}\)
zad.2
Niech \(\displaystyle{ \lambda = \frac{ i + \sqrt{3}}{ 2}}\)
Wówczas \(\displaystyle{ Re( \lambda^{53}) =}\) ....... oraz \(\displaystyle{ Im ( \lambda^{53}) =}\) .........

Wypisać konkretne elementy zbioru: \(\displaystyle{ \{0, 1 , -1, \frac{1}{2}, \sqrt{3} \}}\)

khatus pisze:zad. 1 \(\displaystyle{ x^{2}= (1+8i)x + 16 - 4i}\)
zad.2
Niech \(\displaystyle{ \lambda = \frac{ i + \sqrt{3}}{ 2}}\)
Wówczas \(\displaystyle{ Re( \lambda^{53}) =}\) ....... oraz \(\displaystyle{ Im ( \lambda^{53}) =}\) .........

Wypisać konkretne elementy zbioru: \(\displaystyle{ \{0, 1 , -1, \frac{1}{2}, \sqrt{3} \}}\)

2. \(\displaystyle{ \lambda = \frac{ i + \sqrt{3}}{ 2}}\) = \(\displaystyle{ \frac{\sqrt{3}}{2}+\frac{1}{2}i
= cos(\frac{\pi}{6})+isin(\frac{\pi}{6})}\)

\(\displaystyle{ (cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))^{53}= cos(\frac{\pi *53}{6})+isin(\frac{\pi *53}{6})
=cos(\frac{\pi *5}{6})+isin(\frac{\pi *5}{6}) = -\frac{\sqrt{3}}{2} + i\frac{1}{2}}\)