Graphing Regions and Curves in the Complex Plane

Problem Statement

One of the most visual and rewarding topics in HSC Extension 2 is graphing complex loci. An equation involving a complex variable $z$ does not just represent a single number; it represents a curve or a region in the complex plane.

By interpreting equations geometrically using the modulus (distance) and the argument (angle), we can easily graph lines, rays, circles, and intersecting regions. Often, deep knowledge of classical circle geometry theorems (like the angle in a semicircle) is required to sketch these curves accurately.

Consider a complex number $z = x + iy$. We are given two conditions that $z$ must satisfy simultaneously:

1. $|z - 2i| \le 3$
2. $0 \le \arg(z + 1 - 2i) \le \frac{\pi}{4}$

(a) Describe the geometric meaning of condition 1 and sketch the region it represents on an Argand diagram.

(b) Describe the geometric meaning of condition 2 and sketch the region it represents on the same Argand diagram.

(c) Find the exact Cartesian coordinates $(x, y)$ of the point(s) of intersection between the boundaries of these two regions in the first quadrant.

Hints

• Part (a): The form $|z - (a+ib)| \le R$ represents a solid disc (a circle and its interior). Identify the centre point $a+ib$ and the radius $R$.
• Part (b): The argument $\arg(z - z_0)$ represents the angle formed by a ray starting from the point $z_0$ relative to the positive real horizontal axis. Rewrite the condition as $\arg(z - (-1 + 2i))$. This represents a wedge-shaped region between two rays.
• Part (c): To find the intersection of the boundaries, you must convert the boundary equations into Cartesian form.
  • Boundary 1 is the circle $|z - 2i| = 3$. Convert this to $x^2 + (y-2)^2 = 9$.
  • Boundary 2 is the ray $\arg(z + 1 - 2i) = \frac{\pi}{4}$. Convert this to a Cartesian line by taking the tangent of both sides: $\frac{y-2}{x+1} = \tan(\frac{\pi}{4}) = 1$, with the restriction $x > -1$. Solve these equations simultaneously.

Solutions

Part (a): The Circular Region

1. The condition is $|z - 2i| \le 3$.
2. Rewrite this in the standard locus form: $|z - (0 + 2i)| \le 3$.
3. Geometrically, this represents all points $z$ whose distance from the centre point $(0, 2i)$ is less than or equal to 3.
4. This is a closed circular disc with centre at $C(0, 2)$ and a radius of $R = 3$.
5. Sketching: Draw a circle centred at $(0, 2)$ passing through $(0, 5)$ and $(0, -1)$. Shade the inside.

Part (b): The Angular Wedge

1. The condition is $0 \le \arg(z + 1 - 2i) \le \frac{\pi}{4}$.
2. Rewrite the term inside the argument to identify the starting point: $\arg(z - (-1 + 2i))$.
3. The origin of these rays is the point $P(-1, 2i)$.
4. The region is bounded by two rays starting from $P(-1, 2)$:
   • A ray with an angle of $0$ (horizontal, pointing to the right).
   • A ray with an angle of $\frac{\pi}{4}$ (a $45^\circ$ angle pointing up and right).
5. Sketching: From $(-1, 2)$, draw a horizontal dashed line to the right. Then draw a solid ray at a $45^\circ$ angle. Shade the wedge between them.

Part (c): Finding the Intersection

1. We want the intersection of the boundaries in the first quadrant.
   • Boundary of the circle: $|x + iy - 2i| = 3 \implies x^2 + (y-2)^2 = 9$ (Equation 1)
   • Boundary of the upper ray: $\arg(x+1 + i(y-2)) = \frac{\pi}{4}$. This means $\tan^{-1} \left( \frac{y-2}{x+1} \right) = \frac{\pi}{4}$. $\frac{y-2}{x+1} = \tan(\frac{\pi}{4}) = 1$. Therefore, $y - 2 = x + 1 \implies y = x + 3$ (Equation 2). (Note: Since it's a ray pointing right, we also have the restriction $x > -1$).
2. Substitute Equation 2 into Equation 1: $x^2 + ((x + 3) - 2)^2 = 9$ $x^2 + (x + 1)^2 = 9$
3. Expand and solve for $x$: $x^2 + x^2 + 2x + 1 = 9$ $2x^2 + 2x - 8 = 0$ $x^2 + x - 4 = 0$
4. Use the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2} = \frac{-1 \pm \sqrt{17}}{2}$
5. Since we are looking for the intersection in the first quadrant, $x$ must be positive. Therefore, we take the positive root: $x = \frac{-1 + \sqrt{17}}{2}$
6. Find the corresponding $y$ value using $y = x + 3$: $y = \frac{-1 + \sqrt{17}}{2} + 3 = \frac{-1 + \sqrt{17} + 6}{2} = \frac{5 + \sqrt{17}}{2}$
7. The exact Cartesian coordinates of the intersection point are $\left( \frac{-1 + \sqrt{17}}{2}, \frac{5 + \sqrt{17}}{2} \right)$.

Takeaways

• Geometric Translation: Complex loci problems are essentially translation exercises. You are translating algebraic statements ($|z-a|=r$, $\arg(z-b)=\theta$) into geometric shapes (circles, rays).
• Cartesian Fallback: If you ever get stuck interpreting a complex locus, replace $z$ with $x + iy$ and expand algebraically. This will inevitably yield a standard non-linear Cartesian equation (like a circle or a parabola) that you already know how to graph.
• Circle Geometry: More advanced loci (like $\arg(\frac{z-a}{z-b}) = \theta$) directly rely on circle geometry theorems, specifically the theorem that angles subtended by the same arc at the circumference are equal, which forms circular arcs on the Argand plane.

Further Readings

• HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
• HSC Geometry: https://vumaths.com/booklets/hsc-collections/
• HSC Functions: https://vumaths.com/booklets/hsc-functions/

Connect with me

Graphing complex loci is one of the most aesthetically pleasing parts of the syllabus! To see more examples of these beautiful graphs, head over to Vu's Maths Hub for my complete Extension 2 booklets. I draw these out step-by-step on my YouTube channel. Follow me on Instagram to see some cool mathematical visualisations, and subscribe to my Substack for deeper theoretical dives.