Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 05 (a)/Solution 1

$z=-1/2+i{\sqrt {3}}/2$:

$a=-1/2,b={\sqrt {3}}/2$, it has negative x-value and positive y-value so it should be in the second quadrant. Suppose the angle between $z$ and $x$-axis is $\theta$, we have $\tan \theta ={\frac {b}{a}}=-{\sqrt {3}}$.

It's known $\tan x=-\tan(\pi -x)$ and $\tan(\pi /3)={\sqrt {3}}$, $\theta \in [90^{\circ },180^{\circ }]$. Thus the angle between $z$ and $x$-axis is ${\frac {2\pi }{3}}$