Complex Arguments as Rays, Sectors, and Angle Loci

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Hello! Last lesson, you read $∣ z - a ∣$ as a distance from a fixed point. An argument condition asks a different geometric question: in what direction is the point $z$ from that fixed point? A specified direction produces a ray; a range of directions produces a sector. When the expression is a quotient of two complex differences, the condition can instead describe an angle locus.

The goal today is to recognise which of those pictures a condition calls for, sketch it accurately, and justify the parts of the picture that are easy to miss—especially excluded points and the correct arc of a circle.

A fixed argument gives a ray

Think of $z - a$ as the displacement from the fixed point $a$ to the variable point $z$ . If

ar g (z - a) = θ,

that displacement must point at angle $θ$ to the positive real direction. Its length is free to vary, so the locus is a ray starting at $a$ , not a whole line. More precisely, its points have the form

z = a + r (cos θ + i sin θ), r > 0.

Why $r > 0$ , rather than $r \geq 0$ ? At $z = a$ , the expression $z - a$ is zero, and $ar g 0$ is undefined. Mark the starting point with an open circle.

Read the ray diagrams in 8.2 Argand Diagrams from Save My Exams. They make the distinction between a ray and its excluded starting point particularly clear.

8.2 Argand Diagrams | CIE A Level Maths

Read the explanation of argument loci to see how a fixed direction is drawn from either the origin or a translated point.

In “How do I sketch the locus of arg (z−a) = α on an Argand diagram?” (printed pp. 23–24), read the ray discussion. Begin with the explanation of why the origin is excluded and continue through the example starting at 1+5i. Focus on where the ray starts, which direction it extends, and why its starting point remains open.

For example,

ar g (z - 2 + 3 i) = - \frac{π}{4}

is easier to see when written as $ar g (z - (2 - 3 i)) = - π /4$ . Start at $2 - 3 i$ , then draw a ray down and to the right at $4 5^{\circ}$ below the positive real direction. Do not include $2 - 3 i$ .

If exact coordinates are useful, let $z = x + i y$ . Every point on this ray can be written

z = 2 - 3 i + t (1 - i), t > 0.

Thus $x = 2 + t$ and $y = - 3 - t$ , giving

x + y = - 1, x > 2.

The equation $x + y = - 1$ alone describes an entire line. The restriction $x > 2$ retains the correct half-line and excludes its starting point.

We use the principal argument convention $- π < ar g w \leq π$ . A direction can be described by many angles differing by $2 π$ , but only one is its principal argument. On that convention, an equation such as $ar g (z - a) = 4 π /3$ has no solutions; the ray in that geometric direction is specified by $ar g (z - a) = - 2 π /3$ .

An interval of arguments gives a sector

Now suppose the direction is allowed to vary while the starting point stays fixed. For instance,

- \frac{π}{4} \leq ar g (z - 2 i) < \frac{π}{3} .

All relevant directions are measured from $2 i$ . The locus is the unbounded wedge between the rays at $- π /4$ and $π /3$ :

Include the ray at $- π /4$ , apart from its starting point, because the lower inequality allows equality.
Exclude the ray at $π /3$ , because the upper inequality is strict.
Exclude $2 i$ , even though the first boundary is included: the argument of $z - 2 i$ is undefined there.

This is a sector, not just its two boundary rays. Every direction between the bounds is permitted, and every positive distance from the vertex is permitted.

The inequality diagrams in the same notes show how to shade that region. There is a wording slip in the notes: directions are measured from the fixed point $a$ , not from the angle $α$ .

8.2 Argand Diagrams | CIE A Level Maths

Read the sector explanation to connect an interval of allowable directions with a shaded region.

In “How do I sketch inequalities such as α < arg (z −a) < β on an Argand diagram?” (printed p. 31), read the sector explanation, ending after the instruction to mark the vertex with an open circle. The illustrated bounds are strict; compare them with the mixed inclusive and strict bounds in the example above.

The principal-argument range also matters when a sector crosses the negative real direction. Angles just below $π$ and just above $- π$ point almost the same way, but they lie on opposite sides of the numerical break in principal arguments. Such a sector may need to be described using two argument intervals, rather than an inequality whose lower bound is numerically greater than its upper bound. Sketch the directions first; then check that your inequalities select them.

A quotient specifies an angle at the variable point

A new shape appears when two displacements occur in a quotient:

ar g (\frac{z - a}{z - b}) = θ .

For $z \neq = a, b$ , the quotient’s argument is the directed angle from $z - b$ to $z - a$ , interpreted as a principal angle. Reversing both displacements shows the same directed angle between the rays from the variable point $z$ towards the fixed points $b$ and $a$ . Here the angle’s vertex moves with $z$ ; it is not a ray from either fixed point.

The geometric idea behind this kind of locus is worth seeing before doing algebra. In the following short activity, two fixed points stay on the arms of an angle while its vertex moves.

3 COMPLEX NUMBERS

Read Activity 28 to see why keeping the angle at a moving point fixed can place that point on an arc of a circle.

In Chapter 3, Section 3.12, “Loci in the complex plane” (pp. 84–85), find Activity 28 immediately after Activity 27. Read Activity 28, stopping before the worked example beginning “The point P represents z in the complex plane.” Focus on the roles of the two fixed points and the moving vertex P.

Let us make the arc selection precise with an example:

ar g (\frac{z - 1}{z + 1}) = \frac{π}{2} .

The fixed points are $1$ and $- 1$ . Geometrically, the condition requires a right angle at $z$ between the lines to those points, with a specified direction of turning. Thales’ theorem suggests a semicircle with diameter from $- 1$ to $1$ . But which semicircle?

Write $z = x + i y$ and multiply the quotient by the conjugate of its denominator:

\frac{z - 1}{z + 1} = \frac{x ^{2} + y ^{2} - 1 + 2 i y}{( x + 1 ) ^{2} + y ^{2}}, z \neq = - 1.

For this quotient to have argument $π /2$ , it must be purely imaginary and positive. Its real part must therefore vanish, while its imaginary part must be positive:

x^{2} + y^{2} = 1, y > 0.

So the locus is the upper half of the unit circle, without either endpoint. The circle equation establishes the possible circumference; the sign $y > 0$ selects the correct arc. As a quick check, $z = i$ gives $(i - 1) / (i + 1) = i$ , whose argument is $π /2$ . At $z = 1$ , the quotient is zero and has no argument; at $z = - 1$ , the quotient is undefined.

This illustrates a reliable method for an unfamiliar angle-locus question: identify the fixed points and the angle, use geometry to anticipate a circle arc, then check the orientation and exclusions. Taking a tangent of an argument can help find a circle equation, but tangent alone cannot distinguish angles that differ by $π$ . That is why an equation for the whole circle is not yet a complete answer.

There is a related notation trap. In general,

ar g (\frac{z - a}{z - b})

is the principal value of the difference between the two arguments. A condition that explicitly writes $ar g (z - a) - ar g (z - b) = θ$ may require an additional branch check: subtracting two principal values can produce a number outside the principal-argument range. Do not replace one condition with the other without checking the intended angle.

When distance and direction are combined

A modulus condition from the previous lesson can restrict how far the point travels in the directions allowed by an argument condition. Consider

∣ z - (2 + i) ∣ \leq 2, 0 \leq ar g (z - (2 + i)) \leq \frac{π}{2} .

The first condition gives the closed disk of radius $2$ centred at $2 + i$ . The second permits directions from that centre between rightward and upward, including both boundary directions. Their intersection is a quarter-disk: its outer circular arc and two straight edges are included, but its vertex $2 + i$ is excluded because an argument is undefined there.

In a marking-ready sketch, label the fixed points or centre, mark open endpoints and vertices, and use solid or dashed boundaries to show inclusion. These details turn a plausible diagram into an exact locus.

Key takeaways

Read $ar g (z - a)$ as a direction from $a$ : one permitted direction gives a ray, while an interval gives a sector. The vertex $a$ is always excluded. Read an argument of $(z - a) / (z - b)$ as a directed angle at $z$ between two fixed-point directions; a fixed non-straight angle can give an arc, not necessarily an entire circle. Check principal values, signs and endpoints before declaring the locus complete.

Next, you will bring these ideas together with Cartesian and modulus–argument methods, choosing the form that makes an unfamiliar complex-number problem easiest to solve.