Kompleksni broj na 3000: zbir dva stepena

TEKST ZADATKA

Izračunati:

\bigg(\frac{1+i\sqrt3}2\bigg)^{3000}+\bigg(\frac{1-i\sqrt3}2\bigg)^{3000}

REŠENJE ZADATKA

Primeniti formulu za stepen stepena: $(a^m)^n=a^{m\cdot n}$

\bigg(\bigg(\frac{1+i\sqrt3}2\bigg)^3\bigg)^{1000}+\bigg(\bigg(\frac{1-i\sqrt3}2\bigg)^3\bigg)^{1000}

Srediti razlomke.

\bigg(\frac{(1+i\sqrt3)^3}{2^3}\bigg)^{1000}+\bigg(\frac{(1-i\sqrt3)^3}{2^3}\bigg)^{1000} \\ \bigg(\frac{1+3\sqrt3i+3\cdot3i^2+3\sqrt3i^3}8\bigg)^{1000}+\bigg(\frac{1-3\sqrt3i+3\cdot3i^2-3\sqrt3i^3}8\bigg)^{1000} \\ \bigg(\frac{1+3\sqrt3i+9i^2+3\sqrt3i^3}8\bigg)^{1000}+\bigg(\frac{1-3\sqrt3i+9i^2-3\sqrt3i^3}8\bigg)^{1000}

Pošto je $i^2=-1$ i $i^3=-i$ dobija se:

\bigg(\frac{1+3\sqrt3i-9-3\sqrt3i}8\bigg)^{1000}+\bigg(\frac{1-3\sqrt3i-9+3\sqrt3i}8\bigg)^{1000} \\ \bigg(-\frac88\bigg)^{1000}+\bigg(-\frac88\bigg)^{1000} \\ (-1)^{1000}+(-1)^{1000}

Ako negativna baza ima pozitivan eksponent, baza postaje pozitivna.

1^{1000}+1^{1000} \\ 1+1\\ 2

$\big(\frac{1+i\sqrt3}2\big)^{3000}+\big(\frac{1-i\sqrt3}2\big)^{3000}$