I got a question in my textbox I can't seem to find a valid proof for.
Let $z=a+bi\in\mathbb{C}$ be a complex number which got the polar representation $z = r \operatorname{cis}(\theta)$. What will be the range of $\theta$ given that $b < 0 < a$ and $|b| < |a|$?
I know that since $\arctan$ is a cyclic function its results based on modulus $2\pi$. I also know that the default range of $\theta$ is $0 \le \theta < 2\pi$. I also know that since $b < 0$ and $0 < a$ the number has to be in the 4'th quadrant. I can't find, however, how to connect this information to the general range of $\theta$. If I'd get a particular number I'd just calculate $\theta$ using atan2 but seems like I need more generic explanation in this question.
Thanks!
Since $a > 0$ and $b < 0$, the point $a+bi$ lies in the fourth quadrant (southeast quadrant of the coordinate plane). Moreover, $|a| > |b|$ means that $a+bi$ lies above the line $y=-x$, because in quadrant 4 you have $$|a| > |b| \iff a > -b.$$
Thus the point $a+bi$ lies in the wedge between $y=-x$ and the $x$-axis in the fourth quadrant. The range of angles is $\frac{7\pi}{4} < \theta < 2\pi$.