Subsets of the complex plane
Master subsets of the complex plane in Year 11 VCE Specialist Mathematics. Conditions on a complex number — its modulus, argument, real part or imaginary part — carve out circles, discs, lines and regions in the Argand plane. It sits in the Algebra, number and structure area of study of the VCE Mathematics Study Design (VCAA), within the Complex numbers topic of Unit 2.
You will learn to identify each locus at a glance, convert modulus equations into Cartesian form, and sketch the circles and straight lines they describe — core skills for the complex-number and polar-form work that follows in the course.
Theory
Subsets of the complex plane are regions and curves in the Argand plane described by conditions on a complex number \(z=x+iy\). In Year 11 Specialist Mathematics you identify and sketch the sets fixed by straight lines and circles — modulus, argument, and real- or imaginary-part conditions.
Every complex number \(z=x+iy\) is a point \((x,y)\) in the Argand plane. A condition on \(z\) picks out a set of such points — a curve, a line, or a shaded region.
A modulus condition measures distance. \(|z-a|=r\) is the circle of radius \(r\) centred at the point \(a\); \(|z-a|\le r\) fills it in to a disc (boundary plus interior), and \(|z-a|=|z-b|\) is the perpendicular bisector of the segment joining \(a\) and \(b\).
An argument condition fixes direction. \(\arg(z-a)=\theta\) is a ray starting at \(a\) (with \(a\) itself excluded, since \(\arg 0\) is undefined) pointing at angle \(\theta\) to the positive real direction.
A real- or imaginary-part condition gives lines and half-planes. \(\operatorname{Re}(z)=k\) is the vertical line \(x=k\) and \(\operatorname{Im}(z)=k\) is the horizontal line \(y=k\); replacing \(=\) with an inequality gives a half-plane, and several conditions can be intersected to form a smaller region.
For a complex number \(z=x+iy\), the standard loci are:
How to identify and sketch a subset
- Name the form: is it a modulus \(|z-a|\), an argument \(\arg(z-a)\), or a real/imaginary-part condition? That fixes whether the shape is a circle, a ray, or a line.
- Read off the constants: for \(|z-a|=r\) read the centre \(a\) and radius \(r\); for a bisector read the two points \(a\) and \(b\); for a ray read the start \(a\) and angle \(\theta\).
- Convert if needed: put \(z=x+iy\) and square both sides to get a Cartesian equation for a line, or solve simultaneously for an intersection.
- Sketch and mark: draw the boundary, then use a solid line/disc for \(\le,\ge\) or a dashed boundary for strict \(<,>\), and shade the required region.
Compare the equation with the standard circle \(|z-a|=r\):
| \(|z-(4+2i)|\) | \(=\) | \(3\) |
| \(a\) | \(=\) | \(4+2i\) |
| \(r\) | \(=\) | \(3\) |
The centre is the point \(a\), written as coordinates:
| \(\text{centre}\) | \(=\) | \((4,2)\) |
Circle with centre \(4+2i\) and radius \(3\).
Set \(z=x+iy\); the points are \(-1\) and \(5\). Equate the squared moduli:
| \((x+1)^2+y^2\) | \(=\) | \((x-5)^2+y^2\) |
Expand both sides:
| \(x^2+2x+1+y^2\) | \(=\) | \(x^2-10x+25+y^2\) |
Cancel \(x^2\) and \(y^2\), then collect the \(x\) terms:
| \(2x+1\) | \(=\) | \(-10x+25\) |
| \(12x\) | \(=\) | \(24\) |
| \(x\) | \(=\) | \(2\) |
The vertical line \(x=2\).
The radius is the distance from the centre to the origin, \(|a-0|\):
| \(r\) | \(=\) | \(|3+4i|\) |
| \(=\) | \(\sqrt{3^2+4^2}\) | |
| \(=\) | \(\sqrt{25}\) | |
| \(=\) | \(5\) |
The lowest point sits one radius below the centre, at \(\operatorname{Im}(a)-r\):
| \(\operatorname{Im}(z)_{\min}\) | \(=\) | \(4-5\) |
| \(=\) | \(-1\) |
Radius \(5\); smallest \(\operatorname{Im}(z)=-1\).
The bisector of \(3\) and \(3+8i\): set \(z=x+iy\) and equate squared moduli:
| \((x-3)^2+y^2\) | \(=\) | \((x-3)^2+(y-8)^2\) |
| \(y^2\) | \(=\) | \(y^2-16y+64\) |
| \(16y\) | \(=\) | \(64\) |
| \(y\) | \(=\) | \(4\) |
Substitute \(y=4\) into the circle \(x^2+y^2=25\):
| \(x^2+4^2\) | \(=\) | \(25\) |
| \(x^2\) | \(=\) | \(9\) |
| \(x\) | \(=\) | \(\pm 3\) |
The points \((3,4)\) and \((-3,4)\).
Common pitfalls
Frequently asked questions
What shape is |z - a| = r in the complex plane?
It is a circle. The centre is the point \(a\) and the radius is \(r\), because \(|z-a|\) is the distance from \(z\) to \(a\).
What does |z - a| = |z - b| represent?
The set of points equidistant from \(a\) and \(b\): the perpendicular bisector of the segment joining them. Set \(z=x+iy\) and square both sides to get its Cartesian equation.
Why is the origin excluded from arg(z) = theta?
Because \(\arg 0\) is undefined — the zero complex number has no direction — so \(\arg(z-a)=\theta\) is a ray from \(a\) with the starting point \(a\) left out.
How do I sketch Re(z) > k or Im(z) < k?
Draw the boundary line \(x=k\) (for \(\operatorname{Re}\)) or \(y=k\) (for \(\operatorname{Im}\)) dashed, then shade the half-plane on the required side.
What is the difference between a circle and a disc here?
The equation \(|z-a|=r\) is only the circle (the boundary). The inequality \(|z-a|\le r\) is the disc: the boundary together with every point inside it.
How do I find where a line meets a circle in the Argand plane?
Convert both loci to Cartesian equations, then solve them simultaneously — substitute the line into the circle and solve the resulting equation for the coordinates.