Mathematics 9709 · Pure Mathematics 3

Loci in the complex plane

A locus is the shape z traces out when it follows a distance rule or an angle rule.
Read the rule as a plain sentence about distance: $|z-a|$ means “how far $z$ is from the point $a$ ”.
A distance rule gives a circle or a line. An angle rule gives a ray. Swap $=$ for $\le$ and the shape becomes a region to shade.

Put it in words first, then sketch: $|z-a|$ is how far z is from a; $\arg(z-a)$ is the direction from a to z. Every P3 locus is one of three shapes made from those two ideas.

Three rules, three shapes

$|z-a|=r$ : a circle, centre $a$ , radius $r$ (all the points a fixed distance from $a$ ).
$|z-a|=|z-b|$ : the perpendicular bisector of $a$ and $b$ (the same distance from each).
$\arg(z-a)=\theta$ : a half-line (ray) from $a$ at angle $\theta$ (the point $a$ itself is left out).

Check the sign: $|z+5|$ is really $|z-(-5)|$ , so the centre is $-5$ , not $+5$ .

|z − a| = r is a circle with centre a and radius r (here centre 2 + i, radius 2).

Inequalities: shade the region

Swap $=$ for $\le$ and each boundary becomes a region:

$|z-a|\le r$ : the disc, everything inside the circle.
$|z-a|\le|z-b|$ : the half-plane closer to $a$ than to $b$ (one side of the bisector).
$\alpha\le\arg(z-a)\le\beta$ : the wedge of angles between the two rays.

When a question gives two inequalities, shade only the part where they overlap.
Draw a solid boundary for $\le$ (boundary counts) and a dashed one for a strict $<$ (boundary does not count).

Why |z − a| = |z − b| is a straight line, not a curve

“The same distance from two fixed points” is exactly what a perpendicular bisector means (the same idea from GCSE). In algebra, write $z=x+iy$ , square both sides, and the $x^2$ and $y^2$ terms cancel, leaving a straight-line equation. That cancelling is the clue: one modulus equal to a number gives a circle; one modulus equal to another gives a line.

Worked examples

Worked example

Perpendicular bisector: find the locus of $|z-(-4+i)|=|z-5i|$

Equidistant from two points is the perpendicular bisector. Put $z=x+iy$ and square both sides to clear the roots: $(x+4)^2+(y-1)^2=x^2+(y-5)^2$ .
Expand — the $x^2$ and $y^2$ cancel: $8x+16-2y+1=-10y+25$ , i.e. $8x+8y=8$ .
Divide by 8: the locus is the line $x+y=1$ (it passes through the midpoint $(-2,3)$ ).

Answer

x+y=1

Worked example

Real question: shade $|z-4-2i|\le3$ and $|z|\ge|10-z|$ , then find the greatest $\arg z$

9709/32 Feb/March 2024 Q5. A disc with one side cut off by a half-plane.

First inequality: the disc, centre $4+2i$ , radius 3 (solid boundary).
Second: $|z|\ge|10-z|$ says $z$ is nearer to 10 than to 0 — the half-plane on the $10$ side of the bisector of $0$ and $10$ , namely $\operatorname{Re}z\ge5$ .
For part (b), the greatest $\arg z$ is reached where the line $x=5$ meets the top of the circle, at $z=5+(2+2\sqrt2)\,i$ : $\arg z=\tan^{-1}\dfrac{2+2\sqrt2}{5}$ .

Shade the part of the disc (centre 4 + 2i, radius 3) on or right of Re z = 5; the greatest arg z sits at the top of that slice.

Answer

\text{greatest }\arg z=0.768\ \text{rad}

Where the marks go

Common mistake

\arg(z-a)=\theta

is a ray from

a

(not from O), and

a

itself is left out (open dot). Check the sign of the centre:

|z+5|

has centre

-5

Common mistake

A strict inequality gets a dashed boundary;

\le

gets a solid one. With two inequalities, shade only the overlap — not each region on its own. For greatest/least value parts, the answer is almost always where the boundaries meet, so find those crossing points. (9709/32 Feb/March 2024 Q5)

Now you try

On an Argand diagram, shade the region where $|z-1+2i|\le|z|$ and $|z-2|\le1$ . (9709/33 Oct-Nov 2023 Q2)

Your turn— tap to reveal the worked answer (9709/33/O/N/23 Q2)

$|z-1+2i|\le|z|$ is the half-plane closer to $1-2i$ than to O (the bisector of those two points, solid; in algebra $-2x+4y+5\le0$ ). $|z-2|\le1$ is the disc with centre $2$ and radius 1.

Answer

\text{the overlap region}

Shade the part of the disc $|z-2|\le1$ that sits on the $1-2i$ side of the bisector.

PreviousSolving equations with complex roots

Loci in the complex plane

A locus is the shape z traces out when it follows a distance rule or an angle rule.
Read the rule as a plain sentence about distance: $|z-a|$ means “how far $z$ is from the point $a$ ”.
A distance rule gives a circle or a line. An angle rule gives a ray. Swap $=$ for $\le$ and the shape becomes a region to shade.

Put it in words first, then sketch: $|z-a|$ is how far z is from a; $\arg(z-a)$ is the direction from a to z. Every P3 locus is one of three shapes made from those two ideas.

Three rules, three shapes

$|z-a|=r$ : a circle, centre $a$ , radius $r$ (all the points a fixed distance from $a$ ).
$|z-a|=|z-b|$ : the perpendicular bisector of $a$ and $b$ (the same distance from each).
$\arg(z-a)=\theta$ : a half-line (ray) from $a$ at angle $\theta$ (the point $a$ itself is left out).

Check the sign: $|z+5|$ is really $|z-(-5)|$ , so the centre is $-5$ , not $+5$ .

|z − a| = r is a circle with centre a and radius r (here centre 2 + i, radius 2).

Inequalities: shade the region

Swap $=$ for $\le$ and each boundary becomes a region:

$|z-a|\le r$ : the disc, everything inside the circle.
$|z-a|\le|z-b|$ : the half-plane closer to $a$ than to $b$ (one side of the bisector).
$\alpha\le\arg(z-a)\le\beta$ : the wedge of angles between the two rays.

When a question gives two inequalities, shade only the part where they overlap.
Draw a solid boundary for $\le$ (boundary counts) and a dashed one for a strict $<$ (boundary does not count).

Why |z − a| = |z − b| is a straight line, not a curve

Worked examples

Worked example

Perpendicular bisector: find the locus of $|z-(-4+i)|=|z-5i|$

Equidistant from two points is the perpendicular bisector. Put $z=x+iy$ and square both sides to clear the roots: $(x+4)^2+(y-1)^2=x^2+(y-5)^2$ .
Expand — the $x^2$ and $y^2$ cancel: $8x+16-2y+1=-10y+25$ , i.e. $8x+8y=8$ .
Divide by 8: the locus is the line $x+y=1$ (it passes through the midpoint $(-2,3)$ ).

Answer

x+y=1

Worked example

Real question: shade $|z-4-2i|\le3$ and $|z|\ge|10-z|$ , then find the greatest $\arg z$

9709/32 Feb/March 2024 Q5. A disc with one side cut off by a half-plane.

First inequality: the disc, centre $4+2i$ , radius 3 (solid boundary).
Second: $|z|\ge|10-z|$ says $z$ is nearer to 10 than to 0 — the half-plane on the $10$ side of the bisector of $0$ and $10$ , namely $\operatorname{Re}z\ge5$ .
For part (b), the greatest $\arg z$ is reached where the line $x=5$ meets the top of the circle, at $z=5+(2+2\sqrt2)\,i$ : $\arg z=\tan^{-1}\dfrac{2+2\sqrt2}{5}$ .

Shade the part of the disc (centre 4 + 2i, radius 3) on or right of Re z = 5; the greatest arg z sits at the top of that slice.

Answer

\text{greatest }\arg z=0.768\ \text{rad}

Where the marks go

Common mistake

\arg(z-a)=\theta

is a ray from

a

(not from O), and

a

itself is left out (open dot). Check the sign of the centre:

|z+5|

has centre

-5

Common mistake

A strict inequality gets a dashed boundary;

\le

Now you try

On an Argand diagram, shade the region where $|z-1+2i|\le|z|$ and $|z-2|\le1$ . (9709/33 Oct-Nov 2023 Q2)

Your turn— tap to reveal the worked answer (9709/33/O/N/23 Q2)

$|z-1+2i|\le|z|$ is the half-plane closer to $1-2i$ than to O (the bisector of those two points, solid; in algebra $-2x+4y+5\le0$ ). $|z-2|\le1$ is the disc with centre $2$ and radius 1.

Answer

\text{the overlap region}

Shade the part of the disc $|z-2|\le1$ that sits on the $1-2i$ side of the bisector.