Quick questions on Subsets of the complex plane: VCE Specialist Mathematics Unit 3

$|z - (1 + i)| \le 2$ is the closed disc of radius $2$ centred at the point $(1, 1)$. The boundary circle is drawn solid because the inequality is $\le$.

$\arg(z - (1 + i))$ is the direction from the centre $(1, 1)$ to $z$. Requiring it between $0$ and $\frac{\pi}{2}$ selects the directions from due east round to due north, that is, the upper-right quarter measured from $(1, 1)$. This is the quarter-disc sector.

The region is the quarter of the disc lying up and to the right of the centre: bounded below by the horizontal ray $\arg = 0$ (pointing right from $(1,1)$), on the left by the vertical ray $\arg = \frac{\pi}{2}$ (pointing up), and on the outside by the arc of the circle of radius $2$.

$|z - z_0| = r$ is centred at $z_0$, so $|z + 2 - i| = 3$ is centred at $-2 + i$, not $2 - i$.

Describe the locus $|z - 4i| = 5$. [2 marks]

Sketch $\arg(z) = \frac{3\pi}{4}$. [2 marks]

Describe the region $1 \le |z| \le 2$. [2 marks]