783.

Algebarski oblik

TEKST ZADATKA

Izračunati:

(1+i7)4+(1i7)4(1+i3)4+(1i3)4\frac{(1+i\sqrt7)^4+(1-i\sqrt7)^4}{(-1+i\sqrt3)^4+(-1-i\sqrt3)^4}

REŠENJE ZADATKA

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

((1+i7)2)2+((1i7)2)2((1+i3)2)2+((1i3)2)2(1+27i+7i2)2+(127i+7i2)2(3i223i+1)2+(1+23i+3i2)2\frac{((1+i\sqrt7)^2)^2+((1-i\sqrt7)^2)^2}{((-1+i\sqrt3)^2)^2+((-1-i\sqrt3)^2)^2} \\ \frac{(1+2\sqrt7i+7i^2)^2+(1-2\sqrt7i+7i^2)^2}{(3i^2-2\sqrt3i+1)^2+(1+2\sqrt3i+3i^2)^2}

Pošto je i2=1i^2=-1 dobija se:

(1+27i7)2+(127i7)2(323i+1)2+(1+23i3)2(27i6)2+(27i6)2(223i)2+(23i2)228i2247i+36+28i2+247i+364+83i+12i2+12i283i+456i2+728+24i256+7282416161\frac{(1+2\sqrt7i-7)^2+(1-2\sqrt7i-7)^2}{(-3-2\sqrt3i+1)^2+(1+2\sqrt3i-3)^2}\\ \frac{(2\sqrt7i-6)^2+(-2\sqrt7i-6)^2}{(-2-2\sqrt3i)^2+(2\sqrt3i-2)^2} \\ \frac{28i^2-24\sqrt7i+36+28i^2+24\sqrt7i+36}{4+8\sqrt3i+12i^2+12i^2-8\sqrt3i+4} \\ \frac{56i^2+72}{8+24i^2} \\ \frac{-56+72}{8-24}\\ \frac{16}{-16}\\ -1

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