Упр.629 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
1) 5-i/2-3i;
2) i/(1+i)2;
3) 1+ корень 3i/i3;
4) (1+i)(2+i5)/3-i13.
$$\frac{5-i}{2-3i}=\frac{(5-i)(2+3i)}{(2-3i)(2+3i)}=\frac{10+15i-2i-3i^2}{4-9i^2}=\frac{13+13i}{13}=1+i.$$
$$|z|=\sqrt{1^2+1^2}=\sqrt2,\qquad \tg\varphi=\frac{1}{1}=1,\qquad \varphi=\frac{\pi}{4}.$$
$$z=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right).$$
$$\frac{i}{(1+i)^2}=\frac{i}{1+2i+i^2}=\frac{i}{2i}=\frac12.$$
$$|z|=\frac12,\qquad \varphi=0.$$
$$z=\frac12(\cos 0+i\sin 0).$$
$$\frac{1+\sqrt3\,i}{i^3}=\frac{1+\sqrt3\,i}{-i}=(1+\sqrt3\,i)i=i-\sqrt3.$$
$$|z|=\sqrt{(-\sqrt3)^2+1^2}=2,\qquad \tg\varphi=\frac{1}{-\sqrt3}=-\frac{1}{\sqrt3}.$$
Так как число лежит во II четверти, то $$\varphi=\frac{5\pi}{6}.$$
$$z=2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right).$$
$$\frac{(1+i)(2+i^5)}{3-i^{13}}=\frac{(1+i)(2+i)}{3-i}.$$
$$ (1+i)(2+i)=2+3i+i^2=1+3i,$$
$$\frac{1+3i}{3-i}=\frac{(1+3i)(3+i)}{(3-i)(3+i)}=\frac{3+i+9i+3i^2}{9-i^2}=\frac{10i}{10}=i.$$
$$|z|=1,\qquad \varphi=\frac{\pi}{2}.$$
$$z=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}.$$
Ответ
1) $$\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right);$$ 2) $$\frac12(\cos 0+i\sin 0);$$ 3) $$2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right);$$ 4) $$\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}.$$
