Potenciranje kompleksnih brojeva

Koristeći trigonometrijski oblik kompleksnog broja izračunajte:

Rješenje.

a)

Prema zadatku 1.12 a), trigonometrijski oblik od

$\displaystyle z=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right).$

De Moivreova formula [M1, (1.4)]

daje

$\displaystyle z^{10}$	$\displaystyle =\left(\sqrt{2}\right)^{10}\left[\cos\left(10\cdot\frac{\pi}{4}\right)+i\sin\left(10\cdot\frac{\pi}{4}\right)\right]$
	$\displaystyle =2^5\left(\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}\right)$
	$\displaystyle =32\left[\cos\left(2\pi+\frac{\pi}{2}\right)+i\sin\left(2\pi+\frac{\pi}{2}\right)\right]$
	$\displaystyle =32\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=32(0+i)=32i,$

odnosno vrijedi

$\displaystyle (1+i)^{10}=32i.$

b)

Promotrimo kompleksni broj

$\displaystyle z=\frac{1}{2}-\frac{\sqrt{3}}{2}\,i.$

Prema zadatku 1.12 pod (b), trigonometrijski oblik od

$\displaystyle z=1\cdot\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right).$

De Moivreova formula [M1, (1.4)]

daje

$\displaystyle z^{50}$	$\displaystyle =1^{50}\cdot\left[\cos\left(50\cdot\frac{5\pi}{3}\right)+i\sin\left(50\cdot\frac{5\pi}{3}\right)\right]$
	$\displaystyle =\cos\frac{250\pi}{3}+i\sin\frac{250\pi}{3}$
	$\displaystyle =\cos\left(82\pi+\frac{4\pi}{3}\right)+i\sin\left(82\pi+\frac{4\pi}{3}\right)$
	$\displaystyle =\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}$
	$\displaystyle =-\frac{1}{2}-\frac{\sqrt{3}}{2}i.$

Dakle,

$\displaystyle \left(\frac{1}{2}-\frac{\sqrt{3}}{2}\,i\right)^{50}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i.$