Pj11 - Loci on the Argand diagram | OCR A-Level Further Mathematics B (MEI)

Translate a complex equation into a set of points. You will recognise circles, half-lines and perpendicular bisectors, locate them accurately, and state the features needed to describe each locus completely.

A fixed distance gives a circle

Can an equation involving infinitely many complex numbers describe a familiar shape? Consider $|z-1-2i|=3$ . Its answer is a circle, and the expression tells you its centre and radius without expanding a single square.

You already know

On an Argand diagram, $z=x+yi$ is the point $(x,y)$ . The number $z-a$ represents the displacement from $a$ to $z$ , so $|z-a|$ is their distance. The argument of a non-zero displacement gives its direction.

The points $4+2i$ and $1+5i$ both lie $3$ units from $1+2i$ , so both satisfy the opening equation. The point $1+2i$ itself does not: its distance from that centre is $0$ .

Locus

A locus is the set of all points satisfying a stated condition. A locus is the whole set, not just a few examples of points on it. For example, a circle contains every point at its stated distance from its centre.

Distance turns the complex equation into ordinary coordinate geometry.

Circle locus

For a fixed complex number $a=p+qi$ and a real $r>0$ ,

|z-a|=r \quad\Longleftrightarrow\quad (x-p)^2+(y-q)^2=r^2

This is the circle with centre $(p,q)$ and radius $r$ . If $r=0$ , the set is the single point $a$ ; if $r<0$ , it is empty.

Not in the formula booklet: learn it.

Read the centre from $z-a$ , rather than from a sign spotted in isolation. Brackets make the centre unambiguous.

Worked example 1: reading a circle

Question. Describe the locus $|z-1-2i|=3$ , and give its Cartesian equation.

Plan. Rewrite the displacement as $z-(1+2i)$ , then use the distance formula.

\begin{aligned} |z-(1+2i)| &=3 &&\text{identify the fixed point} \\ |(x-1)+(y-2)i| &=3 &&\text{put }z=x+yi \\ (x-1)^2+(y-2)^2 &=9 &&\text{square the distance} \end{aligned}

Why this step? The number subtracted from $z$ is the centre, so the signs inside the displacement are opposite to its coordinates.

Answer. A circle with centre $(1,2)$ and radius $3$ , with equation $(x-1)^2+(y-2)^2=9$ .

Check. $(4,2)$ gives $9+0=9$ , while the proposed centre gives $0$ , as it should.

Change only the centre's real coordinate from $1$ to $-1$ . The plus sign inside the modulus is the clue to that change.

Worked example 2: a plus sign in the displacement

Question. Describe $|z+1-2i|=3$ , and give its Cartesian equation.

Plan. Write $z+1-2i=z-(-1+2i)$ , then read the centre and radius.

\begin{aligned} |z-(-1+2i)| &=3 &&\text{identify the centre} \\ |(x+1)+(y-2)i| &=3 &&\text{write real coordinates} \\ (x+1)^2+(y-2)^2 &=9 &&\text{square the distance} \end{aligned}

Why this step? Moving the centre left changes $x-1$ to $x+1$ ; it does not change the radius.

Answer. A circle with centre $(-1,2)$ , radius $3$ , and equation $(x+1)^2+(y-2)^2=9$ .

Check. The leftmost and rightmost points are $(-4,2)$ and $(2,2)$ , both $3$ units from $(-1,2)$ .

The picture makes the sign change visible. Both circles have the same size; only their centres move.

[DIAGRAM: asset_name: Reading a translated circle; asset_slug: b003_007_circle_loci; caption: The sign inside z minus a determines the centre, while the modulus value determines the radius; description: Two equal-scale Argand panels show circles of radius 3 centred at (1,2) and (-1,2). Each centre is marked and a horizontal radius is drawn and labelled 3. The first corresponds to absolute value z minus 1 minus 2i equals 3; the second to absolute value z plus 1 minus 2i equals 3.]
The sign inside z minus a determines the centre, while the modulus value determines the radius

Watch out

Wrong: $|z+1-2i|=3$ has centre $(1,-2)$ . Right: rewrite it as $|z-(-1+2i)|=3$ , giving centre $(-1,2)$ . The equation describes the circumference only; its interior has smaller distance.

Describe the next circle by its centre, radius and equation. No drawing is needed for this check.

A fixed direction gives a half-line

The condition $\arg(z-a)=\theta$ fixes the direction of the displacement from $a$ to $z$ , but not its length. Start at $a$ and travel any positive distance in that direction. Travelling the other way changes the argument by $\pi$ , so the locus is half a line, not a whole line.

Half-line locus

For fixed $a=p+qi$ and $-\pi<\theta\le\pi$ ,

\arg(z-a)=\theta \quad\Longleftrightarrow\quad z=a+t(\cos\theta+i\sin\theta),\qquad t>0

The endpoint $a$ is excluded, because $z=a$ would make the argument $\arg0$ , which is undefined. The angle is measured from a horizontal ray pointing right from $a$ .

Not in the formula booklet: learn it.

Use an open circle at the excluded endpoint, an arrow to show the direction, and a labelled angle. A Cartesian line equation needs an extra restriction to select the correct half.

Worked example 3: a half-line above its endpoint

Question. Describe $\arg(z-1-i)=\frac{\pi}{4}$ , and express it using Cartesian coordinates.

Plan. Locate the endpoint $1+i$ , then use a direction $\pi/4$ anticlockwise from the positive horizontal.

\begin{aligned} z &=1+i+t\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right),\qquad t>0 &&\text{fixed direction} \\ x-1 &=\frac{t}{\sqrt2},\qquad y-1=\frac{t}{\sqrt2} &&\text{compare coordinates} \\ y &=x,\qquad x>1 &&\text{eliminate }t \end{aligned}

Why this step? $t>0$ makes both displacements positive. Without $x>1$ , the equation $y=x$ would also include the opposite ray.

Answer. The half-line from $(1,1)$ upwards and right at angle $\pi/4$ , excluding $(1,1)$ ; equivalently $y=x$ , $x>1$ .

Check. At $(2,2)$ , the displacement is $1+i$ , with argument $\pi/4$ . At $(0,0)$ , it is $-1-i$ , with argument $-3\pi/4$ , so that point is excluded.

Change only the angle to $-\pi/4$ . The endpoint stays fixed and the ray now points downwards and right.

Worked example 4: a clockwise direction

Question. Describe $\arg(z-1-i)=-\frac{\pi}{4}$ using Cartesian coordinates.

Plan. Keep the endpoint $(1,1)$ and change the sign of the vertical component of the direction.

\begin{aligned} z &=1+i+t\left(\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i\right),\qquad t>0 &&\text{clockwise angle} \\ x-1 &=\frac{t}{\sqrt2},\qquad y-1=-\frac{t}{\sqrt2} &&\text{compare coordinates} \\ y-1 &=-(x-1) &&\text{eliminate }t \\ y &=2-x,\qquad x>1 &&\text{select the correct ray} \end{aligned}

Why this step? The positive horizontal displacement keeps $x>1$ ; changing the angle's sign changes the vertical displacement, not the starting point.

Answer. $y=2-x$ , $x>1$ , with endpoint $(1,1)$ excluded.

Check. The point $(2,0)$ gives displacement $1-i$ , which has argument $-\pi/4$ .

The endpoint is excluded in both pictures. A fixed direction makes no sense for a zero-length displacement.

[DIAGRAM: asset_name: Half-lines start at the subtracted point; asset_slug: b003_007_half_line_loci; caption: The endpoint stays at 1 plus i while changing the angle reverses the vertical direction; description: Two Argand panels show open endpoints at (1,1). One ray points upwards and right at pi over 4; the other downwards and right at minus pi over 4. Dashed horizontal reference rays begin at the endpoint, and angle arcs are drawn there rather than at the origin.]
The endpoint stays at 1 plus i while changing the angle reverses the vertical direction

Exam tip

For a half-line sketch, show the correct endpoint, the direction and the angle to a horizontal reference. A common error is starting at the negative of the required endpoint. Rewrite $z-a$ before drawing.

Move the endpoint for your turn, keeping the clockwise direction.

Watch out

Wrong: $\arg(z-1-i)=-\pi/4$ gives the whole line $y=2-x$ . The opposite ray has a different argument, and the endpoint has no argument. The correct locus is $y=2-x$ , $x>1$ .

Use both a line equation and a directional restriction for the next locus.

Equal distances give a perpendicular bisector

The condition $|z-a|=|z-b|$ says that $z$ is equally far from two fixed points. If $a\ne b$ , those points form a line segment, and its perpendicular bisector is exactly the set of equal-distance points.

There is a useful algebraic surprise here: both distances contain squares, but their equation becomes linear. The $x^2$ terms cancel, and so do the $y^2$ terms.

Equal-distance locus

For distinct fixed points $a=p+qi$ and $b=u+vi$ ,

|z-a|=|z-b| \quad\Longleftrightarrow\quad (x-p)^2+(y-q)^2=(x-u)^2+(y-v)^2

The locus is the complete perpendicular bisector of the segment joining $a$ and $b$ . If $a=b$ , every point satisfies the equation instead.

Not in the formula booklet: learn it.

Squaring is safe here because both moduli are non-negative. It preserves equality and avoids carrying square roots through the calculation.

Worked example 5: a vertical perpendicular bisector

Question. Describe $|z-1|=|z-5|$ , giving its Cartesian equation.

Plan. Identify the fixed points and their midpoint, then confirm the line by equal squared distances.

\begin{aligned} (x-1)^2+y^2 &=(x-5)^2+y^2 &&\text{equal squared distances} \\ -2x+1 &=-10x+25 &&\text{cancel common squares} \\ 8x &=24 \quad\Longrightarrow\quad x=3 &&\text{solve the linear equation} \end{aligned}

Why this step? The two fixed points $(1,0)$ and $(5,0)$ are horizontally aligned, so their perpendicular bisector must be vertical.

Answer. The complete line $x=3$ , the perpendicular bisector of the segment from $(1,0)$ to $(5,0)$ .

Check. Its point $(3,2)$ is distance $\sqrt8$ from each fixed point.

Now change the fixed points so their joining segment is oblique. The equal-distance idea is unchanged; neither coordinate can be dropped.

Worked example 6: an oblique perpendicular bisector

Question. Find the Cartesian equation of $|z-(1+i)|=|z-(5+3i)|$ , and describe the locus.

Plan. Square both distances, cancel the common terms and simplify to a line.

\begin{aligned} (x-1)^2+(y-1)^2 &=(x-5)^2+(y-3)^2 &&\text{equal squared distances} \\ -2x-2y+2 &=-10x-6y+34 &&\text{cancel }x^2+y^2 \\ 8x+4y &=32 &&\text{collect terms} \\ 2x+y &=8 &&\text{simplify} \end{aligned}

Why this step? The cancelled quadratic terms show why this is a line rather than a circle.

Answer. The line $2x+y=8$ , the perpendicular bisector of the segment joining $(1,1)$ to $(5,3)$ .

Check. The midpoint $(3,2)$ satisfies $2x+y=8$ . The segment has gradient $1/2$ , and the line has gradient $-2$ , so they are perpendicular.

Mark the fixed points and midpoint when sketching. The midpoint is a reliable anchor for an oblique line.

[DIAGRAM: asset_name: Equal distances locate perpendicular bisectors; asset_slug: b003_007_bisector_loci; caption: The equal-distance locus passes through the midpoint and is perpendicular to the segment joining the fixed points; description: Two Argand panels show the segment from (1,0) to (5,0) with perpendicular bisector x equals 3, and the segment from (1,1) to (5,3) with bisector 2x plus y equals 8. Each midpoint is marked. A sample point on each bisector is joined to the two fixed points by equal-length dashed segments.]
The equal-distance locus passes through the midpoint and is perpendicular to the segment joining the fixed points

Watch out

Wrong: the equal-distance locus is the line through the two fixed points. Right: it is perpendicular to that line and passes through its midpoint. Usually, the fixed points themselves are not on the locus.

Apply the same cancellation to a translated pair of points.

Quick Recap

$|z-a|$ is distance from the fixed point $a$ ; $\arg(z-a)$ is the direction from $a$ to $z$ .
$|z-a|=r$ , for $r>0$ , is a circle centred at $a$ with radius $r$ .
$\arg(z-a)=\theta$ is a half-line in direction $\theta$ , with endpoint $a$ excluded.
$|z-a|=|z-b|$ , for $a\ne b$ , is the complete perpendicular bisector of the segment joining $a$ and $b$ .

The opening equation $|z-1-2i|=3$ describes every point on the circle with centre $(1,2)$ and radius $3$ .

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Pj11 - Loci on the Argand diagram

A fixed distance gives a circle

A fixed direction gives a half-line

Equal distances give a perpendicular bisector

Quick Recap

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