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 $\frac{2\pi }{3}$