Ответ:
Используем форму \(z=r(\cos\varphi+i\sin\varphi)\), где \(r=|z|=\sqrt{a^2+b^2}\), а \(\varphi\) определяется по четверти комплексной плоскости.
| № | Тригонометрическая форма | № | Тригонометрическая форма |
|---|---|---|---|
| 1а | \(4\sqrt2(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\) | 1б | \(2\sqrt2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})\) |
| 2а | \(\sqrt2(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))\) | 2б | \(2(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})\) |
| 3а | \(\sqrt{193}(\cos\arctan\frac{7}{12}+i\sin\arctan\frac{7}{12})\) | 3б | \(\sqrt{10}(\cos(\pi-\arctan3)+i\sin(\pi-\arctan3))\) |
| 4а | \(3\sqrt{10}(\cos(-\arctan\frac13)+i\sin(-\arctan\frac13))\) | 4б | \(2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})\) |
| 5а | \(\sqrt{10}(\cos(-\arctan3)+i\sin(-\arctan3))\) | 5б | \(2\sqrt2(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\) |
| 6а, 15а, 12б | \(2(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))\) | 6б | \(2(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})\) |
| 7а | \(\frac{2\sqrt{2074}}5(\cos(-\arctan\frac{45}{7})+i\sin(-\arctan\frac{45}{7}))\) | 7б | \(4(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))\) |
| 8а | \(2\sqrt2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})\) | 8б | \(2\sqrt{10}(\cos(\pi+\arctan\frac13)+i\sin(\pi+\arctan\frac13))\) |
| 9а | \(2(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3}))\) | 9б | \(6(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})\) |
| 10а | \(5\sqrt2(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\) | 10б | \(\sqrt{97}(\cos(\pi-\arctan\frac49)+i\sin(\pi-\arctan\frac49))\) |
| 11а | \(2\sqrt{10}(\cos(\pi-\arctan\frac13)+i\sin(\pi-\arctan\frac13))\) | 11б | \(\sqrt{34}(\cos(-\arctan\frac53)+i\sin(-\arctan\frac53))\) |
| 12а | \(2(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})\) | 12б | \(2(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))\) |
| 13а, 13б | \(\sqrt{65}(\cos\arctan8+i\sin\arctan8)\) | 14а, 14б | \(\sqrt{37}(\cos(\pi-\arctan\frac16)+i\sin(\pi-\arctan\frac16))\) |
| 15б | \(\sqrt{13}(\cos(-\arctan\frac23)+i\sin(-\arctan\frac23))\) | — | — |
