776.

(1+i32)3000+(1−i32)3000 \big(\frac{1+i\sqrt3}2\big)^{3000}+\big(\frac{1-i\sqrt3}2\big)^{3000}

TEKST ZADATKA

Izračunati:

(1+i32)3000+(1−i32)3000 \bigg(\frac{1+i\sqrt3}2\bigg)^{3000}+\bigg(\frac{1-i\sqrt3}2\bigg)^{3000}
REŠENJE ZADATKA

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

((1+i32)3)1000+((1−i32)3)1000 \bigg(\bigg(\frac{1+i\sqrt3}2\bigg)^3\bigg)^{1000}+\bigg(\bigg(\frac{1-i\sqrt3}2\bigg)^3\bigg)^{1000}

Srediti razlomke.

((1+i3)323)1000+((1−i3)323)1000(1+33i+3⋅3i2+33i38)1000+(1−33i+3⋅3i2−33i38)1000(1+33i+9i2+33i38)1000+(1−33i+9i2−33i38)1000 \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 i2=−1i^2=-1 i i3=−ii^3=-idobija se:

(1+33i−9−33i8)1000+(1−33i−9+33i8)1000(−88)1000+(−88)1000(−1)1000+(−1)1000 \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.

11000+110001+121^{1000}+1^{1000} \\ 1+1\\ 2