I'd like to ask you about the example below (I have to draw a geometrical interpretation in an argand diagram). Am I doing the right thing?
$$\frac{\pi}{6}\le \arg\frac{z(1+i)}{-1+i} \le \frac{\pi}{3}.$$
First step:
$$\frac{\pi}{6} \le \arg(-iz) \le \frac{\pi}{3}.$$
Then
$$\arg(-iz) = \arg(-i) + \arg(z) = \arg(z) - \frac{\pi}{2} + 2k\pi,$$ where $k$ is an integer For $k = 0$:
$$\frac{\pi}{6} + \frac{\pi}{2} \le \arg(z) \le \frac{\pi}{3} +\frac{\pi}{2}$$
$$\frac{2\pi}{3} \le \arg(z) \le \frac{5\pi}{6}.$$
Is my geometrical interpretation correct? Should I exclude the $(0,0)$ point?
