Subsets of the complex plane
Explore subsets of the complex plane in Year 11 Specialist Mathematics for Queensland (QCAA). Conditions on a complex number — its modulus, argument, real part or imaginary part — carve out circles, discs, lines and regions in the Argand plane.
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 (QCAA, Queensland) 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.