第1章 1.12

(1)

$(2,2)$

$r=\sqrt{2^{2}+2^{2}}=\sqrt{8}=2\sqrt{2} \qquad \theta=\tan^{-1}\frac{2}{2}=\tan^{-1}1=\frac{\pi}{4} \qquad \therefore (2\sqrt{2},\frac{\pi}{4})$

(2)

$(-\sqrt{3},1)$

$r=\sqrt{(-\sqrt{3})^{2}+1^{2}}=\sqrt{4}=2 \qquad \theta=\tan^{-1}\frac{1}{-\sqrt{3}}=\frac{5}{6}\pi \qquad \therefore (2,\frac{5}{6}\pi)$

(3)

$(2,-2\sqrt{3})$

$r=\sqrt{2^{2}+(-2\sqrt{3})^{2}}=\sqrt{16}=4 \qquad \theta=\tan^{-1}\frac{-2\sqrt{3}}{2}=\tan^{-1}(-\sqrt{3})=\frac{5}{3}\pi \qquad \therefore (4,\frac{5}{3}\pi)$

(4)

$(\sqrt{3},1)$

$r=\sqrt{(\sqrt{3})^{2}+1^{2}}=\sqrt{4}=2 \qquad \theta=\tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{6} \qquad \therefore (2,\frac{\pi}{6})$

(5)

$(0,-\sqrt{3})$

$r=\sqrt{0+(-\sqrt{3})^{2}}=\sqrt{3} \qquad \theta=\tan^{-1}\frac{-\sqrt{3}}{0}=\tan^{-1}(-\infty)=-\frac{\pi}{2}\to\frac{3}{2}\pi \qquad \therefore (\sqrt{3},\frac{3}{2}\pi)$

(6)

$(-1,0)$

$r=\sqrt{(-1)^{2}+0}=1 \qquad \theta=\tan^{-1}\frac{0}{-1}=\tan^{-1}0=\pi \qquad \therefore (1,\pi)$