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\(\sqrt{13}(\cos(-\arctan\frac23)+i\sin(-\arctan\frac23))\)——