Answer to Question #349057 in Calculus for Mapula Advice

Question

Write z1=3–√+i and z2=1−i in trigonometric form. Then the argument of the quotient z1/z2is given by

Answer 1

$$r^2=(-\sqrt{3})^2+(1)^2=4$$ $$r\ge0=>r=\sqrt{4}=2$$

Quadrant II

$$\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{\sqrt{3}}$$ $$\theta=\pi+\tan^{-1}(-\dfrac{1}{\sqrt{3}})=\dfrac{5\pi}{6}$$ $$z_1=2(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6})$$

$$r^2=(1)^2+(-1)^2=2$$ $$r\ge0=>r=\sqrt{2}$$

Quadrant IV

$$\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{1}=-1$$ $$\theta=2\pi+\tan^{-1}(-1)=\dfrac{7\pi}{4}$$ $$z_2=\sqrt{2}(\cos\dfrac{7\pi}{4}+i\sin\dfrac{7\pi}{4})$$

$z_1/z_2$

$$\theta_{z_1/z_2}=\dfrac{5\pi}{6}-\dfrac{7\pi}{4}=-\dfrac{11\pi}{12}$$