776.

Algebarski oblik

TEKST ZADATKA

Izračunati:

(1+i32)3000+(1i32)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=amn(a^m)^n=a^{m\cdot n}

((1+i32)3)1000+((1i32)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+((1i3)323)1000(1+33i+33i2+33i38)1000+(133i+33i233i38)1000(1+33i+9i2+33i38)1000+(133i+9i233i38)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+33i933i8)1000+(133i9+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

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