7. Zadatak

Vrijedi:

$$\begin{aligned} z&=-2\cos\left(\frac{\pi}{2}-\frac{5\pi}{8}\right) -2i\sin\left(\frac{\pi}{2}-\frac{13\pi}{8}\right)\\ &=-2\cos\left(-\frac{\pi}{8}\right) -2i\sin\left(-\frac{9\pi}{8}\right)\\ &=-2\cos\left(\frac{\pi}{8}\right) +2i\sin\left(\frac{9\pi}{8}\right)\\ &=2\left(\cos\left(\frac{9\pi}{8}\right) +i\sin\left(\frac{9\pi}{8}\right)\right), \end{aligned}$$ $$\begin{aligned} w&=\sqrt2\left(\cos\frac{5\pi}{16}-i\sin\frac{5\pi}{16}\right)\\ &=\sqrt2\left(\cos\frac{27\pi}{16}+i\sin\frac{27\pi}{16}\right). \end{aligned}$$

Sada je:

$$\begin{aligned} z^4:w^{12} &=\frac{2^4}{(\sqrt2)^{12}} \left( \cos\left(4\cdot\frac{9\pi}{8}-12\cdot\frac{27\pi}{16}\right) +i\sin\left(4\cdot\frac{9\pi}{8}-12\cdot\frac{27\pi}{16}\right) \right)\\ &=\frac14\left(\cos\left(-\frac{63\pi}{4}\right) +i\sin\left(-\frac{63\pi}{4}\right)\right)\\ &=\frac14\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right). \end{aligned}$$

Modul kompleksnog broja $z^4:w^{12}$ je jednak $\dfrac14$, a njegov argument je jednak $\dfrac{\pi}{4}$.