Extreme Arguments of a Complex Disk

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 $x + iy$ .

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.

Solutions

(a) Geometric Description:

The region is a closed disk (circle plus interior) with:

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

Distance from origin to center: $|OC| = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = 4$ .

Since $|OC| = 4 > r = 2$ , the origin lies outside the circle.

(b) Max and Min Arguments:

The argument of the center is:

\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 create a right triangle with:

Hypotenuse: $|OC| = 4$
Opposite side (radius): $r = 2$

The angle $\theta$ from $OC$ to the tangent satisfies:

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

Therefore:

\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 tangent point distance from origin:

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

The point lies on the ray with argument $\pi/3$ :

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

Takeaways

Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
Check edge cases and verify where each assumption is used in the argument.

Further Readings

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

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Sequences: https://vumaths.com/booklets/hsc-sequences/
HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/

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