1. Muammo: Kompleks sonni darajaga ko'tarish formulasi va misollar bilan tushuntirish. 2. Kompleks sonni darajaga ko'tarish formulasi De Moivre teoremasiga asoslanadi: $$\left( r(\cos \theta + i \sin \theta) \right)^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right)$$ Bu yerda $r$ - kompleks sonning modulidir, $\theta$ - argumenti, $n$ - butun son. 3. Misol: $z = 1 + i$ Avvalo, $r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}$ Argument $\theta = \arctan\left( \frac{1}{1} \right) = \frac{\pi}{4}$ 4. $z^3 = r^3 \left( \cos 3\theta + i \sin 3\theta \right) = (\sqrt{2})^3 \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$ 5. Hisoblaymiz: $$(\sqrt{2})^3 = (2^{1/2})^3 = 2^{3/2} = 2 \sqrt{2}$$ $\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$, $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ 6. Shunday qilib, $$z^3 = 2 \sqrt{2} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 2 \sqrt{2} \left( -\frac{\sqrt{2}}{2} \right) + 2 \sqrt{2} \left( i \frac{\sqrt{2}}{2} \right)$$ $$= 2 \sqrt{2} \cdot -\frac{\sqrt{2}}{2} + i \cdot 2 \sqrt{2} \cdot \frac{\sqrt{2}}{2} = -2 + 2i$$ 7. Natija: $z^3 = -2 + 2i$ Bu usul bilan har qanday kompleks sonni darajaga ko'tarish mumkin.