Kompleksni brojevi: količnik četvrtih stepena binoma

TEKST ZADATKA

Izračunati:

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

REŠENJE ZADATKA

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

\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 $i^2=-1$ dobija se:

\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

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