Core Pure · Argand Diagrams · Year 1

Loci in the Argand Diagram

Conditions on |z − a| and arg(z − a) trace out circles, straight lines, and rays — geometry hiding inside algebraic equations.

17 min read 4 worked examples

Circles: |z − a| = r

Theorem 6.1 — circle locus

The locus of points $z$ satisfying $|z-a|=r$ (where $a$ is a fixed complex number and $r>0$) is a circle centred at the point $a$ with radius $r$, since $|z-a|$ is precisely the distance from $z$ to $a$.

An Argand diagram showing a circle centred at a labelled point a with radius r marked as a line from the centre to the edge of the circle

Fig. 1 — |z − a| = r describes every point at exactly distance r from the fixed point a.

Example 1 · Sketching and interpreting a circle locus

Sketch the locus of points satisfying $|z-(3-2i)|=4$, and state its centre and radius.

This is a circle centred at $(3,-2)$ with radius $4$.

Example 2 · Converting a Cartesian equation to a circle locus

Find the Cartesian equation of the circle described by $|z-4+3i|=5$, and hence confirm it passes through the origin.

Rewrite as $|z-(4-3i)|=5$: circle centred at $(4,-3)$, radius $5$. With $z=x+yi$: $|(x-4)+(y+3)i|=5 \Rightarrow (x-4)^2+(y+3)^2=25$.

At the origin: $(0-4)^2+(0+3)^2=16+9=25$ ✓. The origin lies on the circle.

Method 6.1b — combining a circle locus with a second condition

A common question gives a circle locus $|z-a|=r$ together with a second condition, such as $\operatorname{Re}(z)=k$ or $\operatorname{Im}(z)=k$, and asks for the specific value(s) of $z$ satisfying both. Substitute the second condition directly into the Cartesian circle equation $(x-x_1)^2+(y-y_1)^2=r^2$ and solve — this reduces a two-condition locus problem to ordinary algebra.

Example 1b · Circle meeting a coordinate axis

Given that $z$ satisfies $|z-6|=5$, find the values of $z$ that also satisfy $\operatorname{Im}(z)=0$.

The locus is a circle centred at $(6,0)$ with radius $5$, with Cartesian equation $(x-6)^2+y^2=25$.

Setting $\operatorname{Im}(z)=0$ means $y=0$, so $(x-6)^2=25\Rightarrow x-6=\pm5\Rightarrow x=11$ or $x=1$.

The values of $z$ are $z=11$ and $z=1$ (written as complex numbers, not coordinates).

Perpendicular bisectors: |z − a| = |z − b|

Theorem 6.2 — perpendicular bisector locus

The locus of points $z$ satisfying $|z-a|=|z-b|$ is the set of points equidistant from $a$ and $b$ — this is the perpendicular bisector of the line segment joining $a$ and $b$.

An Argand diagram showing two points a and b connected by a dashed segment, with a solid line perpendicular to the segment through its midpoint

Fig. 2 — |z − a| = |z − b| is the perpendicular bisector of the segment joining the two fixed points a and b.

Example 3 · Finding the equation of a perpendicular bisector

Find the Cartesian equation of the locus $|z-1-2i|=|z+3|$.

Let $z=x+yi$. LHS: $|(x-1)+(y-2)i|=\sqrt{(x-1)^2+(y-2)^2}$. RHS: $|(x+3)+yi|=\sqrt{(x+3)^2+y^2}$.

Squaring both sides: $(x-1)^2+(y-2)^2=(x+3)^2+y^2$.

$x^2-2x+1+y^2-4y+4=x^2+6x+9+y^2$

$-2x+1-4y+4=6x+9 \Rightarrow -8x-4y-4=0 \Rightarrow 2x+y+1=0$.

Half-lines: arg(z − a) = θ

Theorem 6.3 — half-line locus

The locus of points $z$ satisfying $\arg(z-a)=\theta$ (a fixed angle) is a half-line (ray) starting at, but not including, the point $a$, extending in the direction of angle $\theta$ from the positive real-axis direction.

An Argand diagram showing a ray starting from an open circle at point a and extending up and to the right, with the angle theta marked from a horizontal dashed reference line at a

Fig. 3 — arg(z − a) = θ traces only the ray leaving a in direction θ; the point a itself is excluded since the argument there is undefined.

Example 4 · Sketching a half-line locus

Sketch the locus of $\arg(z-2+i)=\dfrac{\pi}{4}$, and find the Cartesian equation of the line it lies on.

This is a half-line from the (excluded) point $(2,-1)$, at angle $\dfrac{\pi}{4}$ to the positive real direction.

Direction has gradient $\tan(\pi/4)=1$. Line: $y-(-1)=1(x-2) \Rightarrow y=x-3$, restricted to $x>2$ (since the ray only extends in the direction of increasing angle from the excluded point, matching the first-quadrant-like direction from $(2,-1)$).

Common mistake

A half-line locus is not a full straight line — only sketch the ray in the correct direction from $a$, with an open circle at $a$ itself to show it's excluded. Examiners specifically penalise diagrams that draw a complete line through $a$ in both directions.

Finding a point satisfying two loci at once

Method 6.4 — intersecting a circle and a half-line

When a question gives both a circle locus $|z-a|=r$ and a half-line locus $\arg(z-a)=\theta$ from the same fixed point $a$, the point satisfying both is found geometrically: it's simply the point at distance $r$ from $a$, in the direction $\theta$. Using right-angled trigonometry (or $x=x_1+r\cos\theta,\ y=y_1+r\sin\theta$) avoids solving simultaneous equations entirely.

Example 5 · Intersecting a circle and half-line from the same point

A complex number $z$ satisfies both $|z-1-2i|=6$ and $\arg(z-1-2i)=\dfrac{\pi}{3}$. Find $z$ in the form $a+bi$, giving exact values.

Both loci share the fixed point $(1,2)$. The circle has radius $6$; the half-line leaves $(1,2)$ at angle $\dfrac{\pi}{3}$ to the horizontal, so the required point is $6$ units from $(1,2)$ in that direction.

$x=1+6\cos\dfrac{\pi}{3}=1+6\left(\dfrac12\right)=4$

$y=2+6\sin\dfrac{\pi}{3}=2+6\left(\dfrac{\sqrt3}{2}\right)=2+3\sqrt3$

So $z=4+(2+3\sqrt3)i$.

Watch out — when the two loci don't share a centre

If the circle and half-line are centred at different fixed points, the shortcut above doesn't apply. Instead, either substitute the half-line's Cartesian equation into the circle's Cartesian equation and solve simultaneously, or sketch both loci accurately and use the geometry of the diagram (e.g. isosceles triangles formed by radii) to find the intersection.

Exam-style questions

Question 1 4 marks

Describe fully the locus given by $|z+2-5i|=3$, stating the centre and radius.

Rewrite as $|z-(-2+5i)|=3$ (M1)

Circle centred at $(-2,5)$ with radius $3$. (A1, A1, A1)

Question 2 6 marks

Find the Cartesian equation of the locus $|z-6i|=|z-4|$, and describe it geometrically.

$z=x+yi$: $x^2+(y-6)^2=(x-4)^2+y^2$ (M1, A1)

$x^2+y^2-12y+36=x^2-8x+16+y^2$

$-12y+36=-8x+16 \Rightarrow 8x-12y+20=0 \Rightarrow 2x-3y+5=0$ (M1, A1, A1)

This is the perpendicular bisector of the segment joining $(0,6)$ and $(4,0)$. (B1)

Question 3 6 marks

The locus $L_1$ is given by $|z-3|=|z-3i|$ and the locus $L_2$ is given by $|z|=3$. Find the point(s) of intersection of $L_1$ and $L_2$.

$L_1$: $(x-3)^2+y^2=x^2+(y-3)^2 \Rightarrow -6x+9=-6y+9 \Rightarrow x=y$ (M1, A1)

$L_2$: $x^2+y^2=9$. Substituting $y=x$: $2x^2=9 \Rightarrow x=\pm\dfrac{3}{\sqrt2}$ (M1, A1)

Points: $\left(\dfrac{3}{\sqrt2},\dfrac{3}{\sqrt2}\right)$ and $\left(-\dfrac{3}{\sqrt2},-\dfrac{3}{\sqrt2}\right)$ (A1, A1)

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Modulus-Argument Form

Regions in the Argand Diagram