MHT-CET Maths · Complex Numbers

Locus in the Argand Plane — Circles, Lines and Greatest/Least Modulus

|z − a| is the distance from z to the point a — so |z − a| = r is a circle, |z − a| = |z − b| is a perpendicular bisector, and the greatest and least |z| on a disc are |a| ± r.

All Complex Numbers notes MHT-CET Maths strategy

Why this matters

12 PYQs at 17% HARD — the cheapest page in the chapter once one sentence is fixed: a modulus is a distance. Every locus question is then geometry: a circle from |z − a| = r or from a ratio of distances, a line from equal distances, and the greatest-and-least-modulus stem that is answered by adding and subtracting a radius — the same move as the Circle chapter's extremum question. The one algebraic member is 'Re of a quotient is zero', which is a circle after rationalising.

Concept 1 of 4

|z − a| = r Is a Circle: Centre a, Radius r

Intuition

|z - a|

measures how far

z

is from the point

a

. Keeping that distance fixed at

r

traces a circle.

|z + 1| = 1

|z - (-1)| = 1

: centre

(-1, 0)

, radius

1

Definition

$|z - (x_0 + iy_0)| = r \iff (x - x_0)^2 + (y - y_0)^2 = r^2$ . Read the centre off the sign: $|z + 1|$ is centred at $-1$ .
$\left|\dfrac{z}{1 + i}\right| = 2$ means $|z| = 2|1 + i| = 2\sqrt2$ : a circle centred at the origin of radius $2\sqrt2$ ( $x^2 + y^2 = 8$ ).
A ratio of distances $\left|\dfrac{z + i}{z - i}\right| = \sqrt3$ is also a circle (Apollonius): square and expand, $x^2 + (y + 1)^2 = 3\left[x^2 + (y - 1)^2\right]$ , giving $x^2 + (y - 2)^2 = 3$ — centre $(0, 2)$ , radius $\sqrt3$ .
Ratio equal to $1$ is the exception: that is a line (next concept).

Circle in the Argand plane

|z - a| = r \iff \text{circle, centre } a,\ \text{radius } r \qquad \left|\frac{z - a}{z - b}\right| = k \ne 1 \iff \text{circle (Apollonius)}

Worked example

Describe the locus

|z - 2 - 3i| = 4

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Complex NumbersEASY

If the complex number

z = x+iy

, where

i = \sqrt{-1}

, satisfies the condition

|z+1| = 1

, then

z

lies on

[Q122 · 15th May Shift 2 · 2023]

Reading |z + 1| as centred at +1

|z + 1| = |z - (-1)|

: the centre is

-1

. The option 'centre

(1, 0)

' is built for this sign slip.

Drill 2 more on |z − a| = r is a circle: centre a, radius r

Concept 2 of 4

|z − a| = |z − b| Is the Perpendicular Bisector of ab

Intuition

Points equidistant from two fixed points lie on the perpendicular bisector of the segment joining them. So an equality of two moduli is a straight line — no circle, whatever the algebra looks like.

Definition

$|z - a| = |z - b|$ : the perpendicular bisector of $a$ and $b$ . $|z + 1 - i| = |z - 1 + i|$ is equidistance from $-1 + i$ and $1 - i$ : squaring gives $4x - 4y = 0$ , the line $y = x$ through the origin in quadrants I and III.
$\left|\dfrac{z - 1}{z + 2i}\right| = 1$ and $\left|\dfrac{z}{z - i/3}\right| = 1$ are the same shape — a line — because the ratio is $1$ .
Difference of distances: $|z + 3| - |z - 3| = 6$ with foci $\pm3$ at distance $6$ apart is the degenerate hyperbola — the ray of the real axis with $x \ge 3$ . The option list says 'X-axis'.
Squaring $|z - a| = |z - b|$ always cancels the $x^2$ and $y^2$ terms; if they do not cancel, the two moduli had different coefficients and the locus is a circle.

Line from equal distances

|z - a| = |z - b| \iff \text{perpendicular bisector of } a, b \qquad |z - a| - |z - b| = |a - b| \ \text{ is a ray (degenerate hyperbola)}

Worked example

Find the locus

|z - 2| = |z + 2i|

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Complex NumbersMODERATE

The equation

|z+ 1 -i| = |z- 1 +i|

represents a (where z is a complex number)

[Q129 · 20 April Shift I · 2025]

Calling every modulus locus a circle

When the two moduli have equal weight the squares cancel and the locus is a LINE. 'Circle' is the first option on every such stem for the student who did not square.

Drill 3 more on |z − a| = |z − b| is the perpendicular bisector of ab

Concept 3 of 4

Re of a Quotient Equals Zero: Rationalise, Then Read the Circle

Intuition

\dfrac{z - 1}{2z + 1}

is purely imaginary' means its real part is

0

. Rationalise, and that real part is a quadratic in

x

and

y

with equal

x^2, y^2

coefficients — a circle whose centre and radius are read by completing the square.

Definition

Real part of $\dfrac{z - 1}{2z + 1}$ with $z = x + iy$ : numerator of the real part is $(x - 1)(2x + 1) + 2y^2 = 2x^2 + 2y^2 - x - 1$ . Setting it to $0$ : $\left(x - \frac14\right)^2 + y^2 = \frac{9}{16}$ , radius $\frac34$ .
On $|z| = 1$ : $\operatorname{Re}\dfrac{z - 1}{z + 1} = \dfrac{x^2 + y^2 - 1}{(x + 1)^2 + y^2} = 0$ — the unit circle maps to the imaginary axis.
An integer-point condition can define a finite locus: $z\bar z^3 + \bar z z^3 = 350$ is $|z|^2(z^2 + \bar z^2) = 2(x^2 + y^2)(x^2 - y^2) = 350$ , i.e. $x^4 - y^4 = 175 = 25\times7$ , so $x^2 = 16$ , $y^2 = 9$ : the four points $(\pm4, \pm3)$ form a rectangle of area $8\times6 = 48$ .
Method: rationalise → separate the real part → set to zero → complete the square.

Purely imaginary quotient

\operatorname{Re}\frac{z - 1}{2z + 1} = 0 \iff 2x^2 + 2y^2 - x - 1 = 0 \iff \left(x - \tfrac14\right)^2 + y^2 = \tfrac{9}{16}

Worked example

\dfrac{z - 2}{z + 2}

is purely imaginary, find the locus of

z

Practice this conceptself-check

From the bank · past-year question

Example 3Complex NumbersHARD

\frac{z- 1}{2z+ 1}

is an imaginary number and if it represents a circle then its radius is

[Q140 · 19 April Shift II · 2025]

Reporting the radius squared

\left(x - \frac14\right)^2 + y^2 = \frac{9}{16}

has radius

\frac34

;

\frac{9}{16}

is the first distractor on the list.

Drill 2 more on re of a quotient equals zero: rationalise, then read the circle

Concept 4 of 4

Greatest and Least |z| on a Disc: |a| + r and |a| − r

Intuition

|z - a| \le r

is a disc centred at

a

. The point of the disc nearest the origin is

r

closer than the centre; the farthest is

r

farther. So the extreme values of

|z|

are

|a| - r

and

|a| + r

, and their difference is

2r

whatever the centre.

Definition

On $|z - a| \le r$ (or $= r$ ): $\max|z| = |a| + r$ , $\min|z| = \big||a| - r\big|$ .
$|z - 2 + i| \le 2$ : centre $2 - i$ , $|a| = \sqrt5$ , so greatest $\sqrt5 + 2$ , least $\sqrt5 - 2$ , difference $4$ .
The difference is always $2r$ when the origin is outside the disc — the question can be answered without computing $|a|$ at all.
Same move for $|z - z_1|$ on a disc centred at $a$ : $|z_1 - a| \pm r$ . It is the Circle chapter's 'maximum distance from a point to a circle' in complex dress.

Extreme modulus on a disc

|z - a| \le r:\quad |a| - r \le |z| \le |a| + r \qquad \max|z| - \min|z| = 2r \ (|a| \ge r)

Worked example

|z - 3 - 4i| \le 2

, find the greatest and least values of

|z|

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Complex NumbersMODERATE

|z - 2 + i| \leq 2

, then the difference between the greatest and least value of

|z|

(i = \sqrt{-1})

[Q102 · 13th May Shift 1 · 2024]

Answering 2√5 for the difference

The difference of the extreme moduli is

2r

, the diameter, not

2|a|

. For

|z - 2 + i| \le 2

that is

4

;

2\sqrt5

is the planted wrong answer.

Drill 1 more on greatest and least |z| on a disc: |a| + r and |a| − r

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

|z − a| = r Is a Circle: Centre a, Radius r
Circle in the Argand plane
$|z - a| = r \iff \text{circle, centre } a,\ \text{radius } r \qquad \left|\frac{z - a}{z - b}\right| = k \ne 1 \iff \text{circle (Apollonius)}$
|z − a| = |z − b| Is the Perpendicular Bisector of ab
Line from equal distances
$|z - a| = |z - b| \iff \text{perpendicular bisector of } a, b \qquad |z - a| - |z - b| = |a - b| \ \text{ is a ray (degenerate hyperbola)}$
Re of a Quotient Equals Zero: Rationalise, Then Read the Circle
Purely imaginary quotient
$\operatorname{Re}\frac{z - 1}{2z + 1} = 0 \iff 2x^2 + 2y^2 - x - 1 = 0 \iff \left(x - \tfrac14\right)^2 + y^2 = \tfrac{9}{16}$
Greatest and Least |z| on a Disc: |a| + r and |a| − r
Extreme modulus on a disc
$|z - a| \le r:\quad |a| - r \le |z| \le |a| + r \qquad \max|z| - \min|z| = 2r \ (|a| \ge r)$

Watch out for (4)

Reading |z + 1| as centred at +1
→ |z − a| = r Is a Circle: Centre a, Radius r
Calling every modulus locus a circle
→ |z − a| = |z − b| Is the Perpendicular Bisector of ab
Reporting the radius squared
→ Re of a Quotient Equals Zero: Rationalise, Then Read the Circle
Answering 2√5 for the difference
→ Greatest and Least |z| on a Disc: |a| + r and |a| − r

Drill every past-year question on this subtopic

12 questions from the bank — paginated, with cart and Word-export support.

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Related notes

Modulus and Argument — |z| as a distance, the fact every locus here rests on →