Extreme Arguments of a Complex Disk

This problem dives into the geometric side of complex numbers, a favourite in HSC Extension 2. You’ll use the modulus form to identify a disk, then combine right‑angle trigonometry with the argument of the centre to find the extreme arguments. Mastering this will strengthen your ability to visualise and solve modulus‑argument problems without crunching through messy algebra.

Problem Statement

Consider the region $\mathcal{R}$ in the complex plane defined by $|z - (2\sqrt{3} + 2i)| \leq 2$.

• Describe the region $\mathcal{R}$ geometrically (what shape? center? radius?).
• Find the maximum and minimum values of $\arg(z)$ for $z \in \mathcal{R}$, where $-\pi < \arg(z) \leq \pi$.
• Find the complex number $z$ associated with the maximum argument in part (b) in the form $x + iy$.

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Hints

(a) The inequality $|z - w| \leq r$ represents a disk (filled circle) in the complex plane.

(b) Find $\arg(C)$ where $C = 2\sqrt{3} + 2i$ is the center. The max/min arguments occur at tangent lines from the origin to the circle. Use the right triangle formed by the origin, center, and tangent point.

(c) The point with maximum argument lies on the ray from $O$ with angle $\arg_{\max}$ found in (b), at distance $\sqrt{|OC|^2 - r^2}$ from origin.

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Solutions

(a) Geometric Description:

The inequality $|z - (2\sqrt{3} + 2i)| \le 2$ directly tells us we are dealing with a disk — all points whose distance from the centre is at most $2$. So we read off:

• Center: $C = 2\sqrt{3} + 2i$ (which is $(2\sqrt{3}, 2)$ in Cartesian coordinates)
• Radius: $r = 2$

To later decide whether the disk contains the origin, we calculate the distance from the origin to the centre:

$$|OC| = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = 4.$$

Since $|OC| = 4 > r = 2$, the origin lies outside the circle. This is crucial because if the origin were inside the disk, the argument could take any value, and we wouldn't get a restricted range.

(b) Max and Min Arguments:

Because the origin is outside the disk, the rays that give the maximum and minimum arguments are exactly the two tangent lines from the origin to the circle. To find them we first get the argument of the centre as a reference:

$$\arg(C) = \arctan\left(\frac{2}{2\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.$$

The tangent lines from $O$ to the circle form two congruent right triangles: each has the line $OC$ as hypotenuse, the radius as one leg, and the tangent segment as the other leg. The angle $\theta$ between $OC$ and each tangent ray satisfies

$$\sin \theta = \frac{r}{|OC|} = \frac{2}{4} = \frac{1}{2} \implies \theta = \frac{\pi}{6}.$$

Therefore the extreme arguments are obtained by adding and subtracting this angle from $\arg(C)$:

$$\begin{aligned} \arg_{\max} &= \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \\ \arg_{\min} &= \frac{\pi}{6} - \frac{\pi}{6} = 0 \end{aligned}$$

(c) Complex Number for Max Argument:

The point of tangency for the maximum argument lies on the ray at angle $\pi/3$. Its distance from the origin is the length of the tangent segment, found directly from Pythagoras in the right triangle:

$$|OP| = \sqrt{|OC|^2 - r^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}.$$

Putting this together in polar form and converting to Cartesian coordinates:

$$\begin{aligned} z &= 2\sqrt{3} \cdot e^{i\pi/3} = 2\sqrt{3}\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) \\ &= 2\sqrt{3}\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \\ &= \sqrt{3} + 3i. \end{aligned}$$

Answer: $z = \sqrt{3} + 3i$. $\blacksquare$

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Takeaways

• The extreme arguments of a disk with origin outside occur at the tangent points; the angle between the centre ray and a tangent is $\theta = \arcsin(r/|OC|)$.
• The maximum (or minimum) argument equals $\arg(C) \pm \theta$, and the corresponding point lies at distance $\sqrt{|OC|^2 - r^2}$ along that ray.
• Always check whether the origin is inside or outside the disk — if inside, the argument range is unrestricted ($-\pi < \arg(z) \le \pi$).

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Sequences, HSC Mechanics