Complex Numbers in Coordinate Geometry Complete JEE Notes

Coordinate geometry gives every point two numbers. Complex numbers give it one, and that single change is what makes this topic worth a chapter of its own. A rotation, which costs you a page of trigonometry in Cartesian coordinates, becomes one multiplication. A circle, a line, an ellipse each collapses to a short statement about distances or angles.

JEE has noticed. Across the ten JEE Main 2026 shifts we hold verified solutions for, thirteen questions were built on complex numbers, and nine of those thirteen were pure geometry circles, perpendicular bisectors, an annulus, an ellipse, a maximum-argument problem. Not one of them needed a hard algebraic identity. All of them needed a picture.

The two numbers that describe $z$. The modulus $\lvert z\rvert=\sqrt{x^2+y^2}$ is the distance $OP$. The argument is the angle $OP$ makes with the positive real axis, measured anticlockwise as positive. Because a full turn returns you to the same point, the argument is only defined up to $2n\pi$; the value in $(-\pi,\pi]$ is the principal argument, written $\arg z$.

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The four operations, seen as pictures

Addition places the two vectors nose to tail: $z_1+z_2$ is the fourth vertex of the parallelogram with $O$, $z_1$, $z_2$. Subtraction is the vector from $z_2$ to $z_1$, which is why

$\lvert z_1-z_2\rvert = \text{the distance between the points } z_1 \text{ and } z_2.$

That one line is the most used fact in the chapter.

Multiplication rotates and scales. Writing $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$,

$z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)},\qquad \frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}.$

Moduli multiply, arguments add. Dividing subtracts the argument instead. In particular multiplying by $i=e^{i\pi/2}$ is exactly a quarter-turn anticlockwise, and multiplying by $-1$ is a half-turn.

Example 1. If $z=2+3i$, describe $iz$, $-z$ and $\bar z$ geometrically, and find each.

▸ SHOW SOLUTION

• $iz=i(2+3i)=-3+2i$ the point turned a quarter-turn anticlockwise about the origin.
• $-z=-2-3i$ — turned a half-turn, i.e. reflected in the origin.
• $\bar z=2-3i$ — reflected in the real axis.

Each has the same modulus $\sqrt{13}$, because rotation and reflection preserve distance from the origin.

Rotation — the master tool

Everything geometric in this chapter is downstream of one formula.

The rotation formula. If $A(z_1)$, $B(z_2)$, $C(z_3)$ are three points and $\alpha$ is the angle from $AB$ to $AC$ measured anticlockwise, then $\frac{z_3-z_1}{z_2-z_1}=\frac{\lvert z_3-z_1\rvert}{\lvert z_2-z_1\rvert}\;e^{i\alpha}.$

Why ? $z_2-z_1$ is the vector $AB$ and $z_3-z_1$ is the vector $AC$. Dividing two complex numbers divides their moduli and subtracts their arguments, so the quotient has modulus $AC/AB$ and argument exactly the angle between them. There is nothing more to it.

Example 2 (IIT-JEE 2008). A particle $P$ starts at $z_0=1+2i$. It first moves $5$ units horizontally away from the origin, then $3$ units vertically away from the origin, reaching $z_1$. From $z_1$ it moves $\sqrt2$ units in the direction of $\hat i+\hat j$ and then turns through $\dfrac{\pi}{2}$ anticlockwise on a circle centred at the origin, reaching $z_2$. Find $z_2$.

▸ SHOW SOLUTION

Step 1: The two axis moves. $z_0=1+2i$ has positive real and imaginary parts, so "away from the origin" means the $+x$ and $+y$ directions:

$1+2i \xrightarrow{\;+5\;} 6+2i \xrightarrow{\;+3i\;} z_1=6+5i.$

Step 2: The diagonal move. The unit vector along $\hat i+\hat j$ is $\dfrac{1+i}{\sqrt2}$, so moving $\sqrt2$ units adds $\sqrt2\cdot\dfrac{1+i}{\sqrt2}=1+i$:

$6+5i \longrightarrow 7+6i.$

Step 3: The rotation. Turning through $\dfrac{\pi}{2}$ anticlockwise about the origin means multiplying by $e^{i\pi/2}=i$:

$z_2=i(7+6i)=7i-6=\boxed{-6+7i}$

Notice that the whole question is four one-line moves. Setting up coordinates and rotation matrices would take five times as long.

Example 3. $z_1$ and $z_2$ are two adjacent vertices of a square, taken anticlockwise, with the centre at the origin. Express the other two vertices in terms of $z_1$.

▸ SHOW SOLUTION

The four vertices of a square centred at the origin are equally spaced round a circle, at $\dfrac{\pi}{2}$ apart. Successive anticlockwise vertices are therefore obtained by multiplying by $i$:

$z_1,\quad z_2=iz_1,\quad z_3=i^2z_1=-z_1,\quad z_4=i^3z_1=-iz_1.$

Check. $z_1+z_3=0=z_2+z_4$, so the diagonals bisect each other at the origin; and $\lvert z_1-z_3\rvert=2\lvert z_1\rvert=\lvert z_2-z_4\rvert$, so the diagonals are equal. Both are properties a square must have.

Distance, section formula and collinearity

Because $\lvert z_2-z_1\rvert$ is the distance $P_1P_2$, every result of coordinate geometry transfers unchanged.

Section formula. If $R(z)$ divides the join of $P(z_1)$ and $Q(z_2)$ in the ratio $m:n$, then $z=\frac{mz_2+nz_1}{m+n}\ \ \text{(internally)},\qquad z=\frac{mz_2-nz_1}{m-n}\ \ \text{(externally)}.$ In particular the midpoint is $\dfrac{z_1+z_2}{2}$ and the centroid of $\triangle$ with vertices $z_1,z_2,z_3$ is $\dfrac{z_1+z_2+z_3}{3}$.

Collinearity. $A(z_1)$, $B(z_2)$, $C(z_3)$ are collinear if and only if any of the following holds — they are the same statement three ways:

1. $\dfrac{z_3-z_1}{z_2-z_1}$ is purely real;
2. $\lvert z_1-z_2\rvert+\lvert z_2-z_3\rvert=\lvert z_1-z_3\rvert$ (with $B$ between $A$ and $C$);
3. $\begin{vmatrix} z_1 & \bar z_1 & 1\\ z_2 & \bar z_2 & 1\\ z_3 & \bar z_3 & 1\end{vmatrix}=0$.

Area of a triangle. $\displaystyle \Delta=\frac14\left\lvert \begin{vmatrix} z_1 & \bar z_1 & 1\\ z_2 & \bar z_2 & 1\\ z_3 & \bar z_3 & 1\end{vmatrix}\right\rvert$, which is why the determinant vanishing means collinear.

Example 4 (IIT-JEE 2010). Let $z_1,z_2$ be distinct complex numbers and let $z=(1-t)z_1+tz_2$ for some real $t$ with $0<t<1$. If $\arg(w)$ denotes the principal argument of a non-zero complex number $w$, which of the following is/are true?

(a) $\lvert z-z_1\rvert+\lvert z-z_2\rvert=\lvert z_1-z_2\rvert$   (b) $\arg(z-z_1)=\arg(z_2-z_1)$   (c) $\begin{vmatrix} z-z_1 & \bar z-\bar z_1\\ z_2-z_1 & \bar z_2-\bar z_1\end{vmatrix}=0$   (d) $\arg(z-z_1)=\arg(z_2-z_1)$

▸ SHOW SOLUTION

Step 1: Recognise the parametrisation. $z=(1-t)z_1+tz_2$ is exactly the section formula with $z$ dividing $z_1z_2$ in the ratio $t:(1-t)$. Since $0<t<1$, the point $z$ lies strictly between $z_1$ and $z_2$ on the segment.

Step 2: Test each statement.

• (a) $z$ lies on the segment, so the two part-lengths add to the whole. True.
• (b), (d) $z-z_1=t(z_2-z_1)$ with $t>0$, so the two vectors point the same way and have the same principal argument. True.
• (c) The determinant is $(z-z_1)(\bar z_2-\bar z_1)-(\bar z-\bar z_1)(z_2-z_1)$. Substituting $z-z_1=t(z_2-z_1)$ gives $t\lvert z_2-z_1\rvert^2-t\lvert z_2-z_1\rvert^2=0$. True — it is the collinearity determinant.

Answer: all of them.

Triangles

Two results carry almost all the marks.

Equilateral triangle. $z_1,z_2,z_3$ form an equilateral triangle iff $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1,$ equivalently $\dfrac{1}{z_1-z_2}+\dfrac{1}{z_2-z_3}+\dfrac{1}{z_3-z_1}=0$. If in addition the circumcentre is at the origin, then $z_1+z_2+z_3=0$.

Why the first form is true. Rotating $B$ about $A$ through $\dfrac{\pi}{3}$ lands on $C$, so $\dfrac{z_3-z_1}{z_2-z_1}=e^{i\pi/3}$. The same is true starting from each vertex. Writing $\omega=e^{i\pi/3}$, eliminating between the three relations produces exactly the symmetric identity above.

Example 5. An equilateral triangle has its circumcentre at the origin and circumradius $2$. Find $(AB)^2+(BC)^2+(CA)^2$, and $(PA)^2+(PB)^2+(PC)^2$ for any point $P$ on its incircle.

▸ SHOW SOLUTION

Step 1: Place the vertices. With circumcentre $O$ at the origin, take $z_1=2$, $z_2=2\omega$, $z_3=2\omega^2$ where $\omega=e^{2\pi i/3}$. Then $z_1+z_2+z_3=0$.

Step 2: The side. $\lvert z_1-z_2\rvert=2\sqrt3$, so each side is $a=R\sqrt3=2\sqrt3$ and

$(AB)^2+(BC)^2+(CA)^2=3a^2=3\cdot12=\boxed{36}.$

Step 3: The incircle. For an equilateral triangle the inradius is half the circumradius, so $r=1$ and $P$ satisfies $\lvert P\rvert=1$.

Step 4: Use the centroid identity. For any point $P$,

$\sum \lvert P-z_k\rvert^2=3\lvert P-G\rvert^2+\sum\lvert z_k-G\rvert^2,$

and here the centroid $G$ is the origin. So the sum is $3(1)^2+3(2)^2=3+12=\boxed{15}$, independent of where $P$ sits on the incircle.

Check. Direct expansion with $P=e^{i\theta}$ gives $15$ for every $\theta$ — the $\theta$-terms cancel because $z_1+z_2+z_3=0$.

Triangle centres. With vertices $z_1,z_2,z_3$ and opposite side lengths $a,b,c$:

Centre	Formula
Centroid $G$	$\dfrac{z_1+z_2+z_3}{3}$
Incentre $I$	$\dfrac{az_1+bz_2+cz_3}{a+b+c}$
Circumcentre $O$, orthocentre $H$	collinear with $G$, and $H=3G-2O$

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Quadrilaterals

Four points $z_1,z_2,z_3,z_4$ taken in order form:

Example 6. $P,Q,R,S$ are represented by $4+i$, $1+6i$, $-4+3i$, $-1-2i$ respectively. What is $PQRS$?

▸ SHOW SOLUTION

Step 1: Test the parallelogram condition.

$z_1+z_3=(4+i)+(-4+3i)=4i,\qquad z_2+z_4=(1+6i)+(-1-2i)=4i.$

Equal, so the diagonals bisect each other — it is at least a parallelogram.

Step 2: Adjacent sides.

$\lvert z_2-z_1\rvert=\lvert-3+5i\rvert=\sqrt{34},\qquad \lvert z_4-z_1\rvert=\lvert-5-3i\rvert=\sqrt{34}.$

Equal, so it is a rhombus.

Step 3: Diagonals.

$\lvert z_3-z_1\rvert=\lvert-8+2i\rvert=\sqrt{68},\qquad \lvert z_4-z_2\rvert=\lvert-2-8i\rvert=\sqrt{68}.$

Equal, so it is also a rectangle.

Answer: rhombus and rectangle, i.e. a $\boxed{\text{square}}$.

Straight lines

The perpendicular bisector. $\lvert z-z_1\rvert=\lvert z-z_2\rvert$ says $z$ is equidistant from two fixed points, so the locus is the perpendicular bisector of the segment joining them.

Example 7. Find the locus of $z=x+iy$ satisfying $\lvert z-(2-i)\rvert=\lvert z-(3+i)\rvert$.

▸ SHOW SOLUTION

Step 1: Recognise it. Equal distances from $2-i$ and $3+i$ — the perpendicular bisector of the segment joining $(2,-1)$ and $(3,1)$.

Step 2: Square both sides and expand:

$(x-2)^2+(y+1)^2=(x-3)^2+(y-1)^2.$

Step 3: Cancel $x^2$ and $y^2$.

$-4x+4+2y+1=-6x+9-2y+1\ \Longrightarrow\ 2x+4y-5=0.$

Answer: the line $\boxed{2x+4y-5=0}$.

Check. The midpoint is $\left(\dfrac52,0\right)$, and $2\cdot\dfrac52+0-5=0$. The direction of $PQ$ is $(1,2)$, and the line's normal is $(2,4)\parallel(1,2)$. Both as they must be.

General equation of a line. $\bar a z+a\bar z+b=0$ with $a\in\mathbb C$, $b\in\mathbb R$. Its slope is $-\dfrac{a}{\bar a}$ (as a complex direction). Two lines with $a_1,a_2$ are parallel when $\dfrac{a_1}{\bar a_1}=\dfrac{a_2}{\bar a_2}$ and perpendicular when $\dfrac{a_1}{\bar a_1}+\dfrac{a_2}{\bar a_2}=0$.

Parametric form. $z=(1-t)z_1+tz_2$, $t\in\mathbb R$, is the whole line through $z_1$ and $z_2$; restricting to $0\le t\le1$ gives the segment.

Determinant form. $\begin{vmatrix} z & \bar z & 1\\ z_1 & \bar z_1 & 1\\ z_2 & \bar z_2 & 1\end{vmatrix}=0$.

Circles

The four forms.

1. Centre–radius: $\lvert z-z_0\rvert=r$.
2. General: $z\bar z+\bar a z+a\bar z+b=0$ with $a\in\mathbb C$, $b\in\mathbb R$ — centre $-a$, radius $\sqrt{\lvert a\rvert^2-b}$.
3. Diameter form: if $z_1,z_2$ are ends of a diameter, $(z-z_1)\overline{(z-z_2)}+\overline{(z-z_1)}(z-z_2)=0$.
4. Right-angle form: $\lvert z-z_1\rvert^2+\lvert z-z_2\rvert^2=\lvert z_1-z_2\rvert^2$ — again the circle on $z_1z_2$ as diameter, by Pythagoras.

Nature of the general equation. For $z\bar z+\bar az+a\bar z+b=0$, look at $\lvert a\rvert^2-b$:

Sign	What you get
$\lvert a\rvert^2>b$	a real circle of radius $\sqrt{\lvert a\rvert^2-b}$
$\lvert a\rvert^2=b$	a point circle — the single point $z=-a$
$\lvert a\rvert^2<b$	an imaginary circle — no points at all

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Example 8 (JEE Main 2026, 28 Jan Shift 1). Let $z$ satisfy $\lvert z-6\rvert=5$ and $\lvert z+2-6i\rvert=5$. Find $z$.

▸ SHOW SOLUTION

Step 1: Two circles. Centres $C_1=6$ and $C_2=-2+6i$, both of radius $5$.

Step 2: Distance between centres.

$\lvert C_1-C_2\rvert=\lvert 8-6i\rvert=\sqrt{64+36}=10=5+5.$

The centres are exactly $r_1+r_2$ apart, so the circles touch externally — there is precisely one common point.

Step 3: The point of contact divides $C_1C_2$ in the ratio $r_1:r_2=1:1$, i.e. it is the midpoint:

$z=\frac{6+(-2+6i)}{2}=\frac{4+6i}{2}=\boxed{2+3i}$

Recognising the touching case turns a pair of simultaneous quadratics into one midpoint.

Example 9 (JEE Main 2026, 23 Jan Shift 1). Let $S=\{z:3\le\lvert 2z-3(1+i)\rvert\le7\}$. Find $\displaystyle\min_{z\in S}\left\lvert z+\tfrac12(5+3i)\right\rvert$.

▸ SHOW SOLUTION

Step 1: Normalise the coefficient of $z$. Pull the $2$ out of the modulus:

$\lvert 2z-3(1+i)\rvert=2\left\lvert z-\tfrac32(1+i)\right\rvert,$

so the condition becomes $\dfrac32\le\left\lvert z-\left(\dfrac32+\dfrac32i\right)\right\rvert\le\dfrac72$.

Step 2: Name the region. $S$ is an annulus — the ring between two concentric circles — centred at $C\left(\dfrac32,\dfrac32\right)$ with inner radius $\dfrac32$ and outer radius $\dfrac72$.

Step 3: Name the target. $\left\lvert z+\dfrac12(5+3i)\right\rvert=\left\lvert z-\left(-\dfrac52-\dfrac32i\right)\right\rvert$ is the distance from $z$ to the fixed point $P\left(-\dfrac52,-\dfrac32\right)$.

Step 4: Locate $P$ relative to the ring.

$PC=\sqrt{\left(\tfrac32+\tfrac52\right)^2+\left(\tfrac32+\tfrac32\right)^2}=\sqrt{4^2+3^2}=5.$

Since $5>\dfrac72$, the point $P$ lies outside the annulus altogether.

Step 5: Minimise. For a point outside, the nearest point of the ring lies on the segment $PC$, on the outer circle:

$\min=PC-\frac72=5-\frac72=\boxed{\frac32}$

Had $P$ fallen inside the hole, the answer would instead have been $\dfrac32-PC$; had it fallen within the ring, the minimum would be $0$. Deciding which of the three cases you are in is the entire question.

Products of distances to points on a circle. If $A_1,\dots,A_n$ are the vertices of a regular $n$-gon inscribed in $|z|=R$, they are exactly the roots of $z^n=R^n$. Hence for any $z$, $\prod_{k=1}^{n}\left(z-A_k\right)=z^{n}-R^{n},$ so a product of $n$ distances collapses to a single modulus. This is the only sensible way to handle "find $PA_1\cdot PA_2\cdots PA_n$".

Example 10 (JEE Advanced 2023). $A_1,A_2,\dots,A_8$ are the vertices of a regular octagon inscribed in a circle of radius $2$, and $P$ is a point on that circle. Find the maximum value of $PA_1\cdot PA_2\cdots PA_8$.

▸ SHOW SOLUTION

Step 1: Put the octagon at the roots of an equation. Place the centre at the origin. The eight vertices are equally spaced on $|z|=2$, so they are precisely the eight solutions of

$z^{8}=2^{8}=256.$

Step 2: Turn the product of distances into one modulus. If $P$ is the point $z$, then $PA_k=\lvert z-A_k\rvert$, and since the $A_k$ are all the roots of $z^8-256$,

$\prod_{k=1}^{8}\left(z-A_k\right)=z^{8}-256 \quad\Longrightarrow\quad PA_1\cdot PA_2\cdots PA_8=\left\lvert z^{8}-256\right\rvert.$

Step 3: Use $\lvert z\rvert=2$. Then $\lvert z^{8}\rvert=2^{8}=256$, so $z^8$ itself runs over the circle of radius $256$ centred at the origin.

Step 4: Maximise. $\left\lvert z^{8}-256\right\rvert$ is the distance from the point $z^8$ to the fixed point $256$, and both lie on (or the latter on) the circle of radius $256$. The greatest such distance is the diameter:

$\left\lvert z^{8}-256\right\rvert\le 256+256=512,$

attained when $z^{8}=-256$, i.e. when $P$ is a midpoint of an arc between two adjacent vertices.

Answer: $\boxed{512}$

Numerical check. Sweeping $P$ round the circle in $4000$ steps, the product peaks at $512.000$ — exactly at the eight arc midpoints.

Loci by ratio — the Apollonius circle

Apollonius. For fixed $z_1\neq z_2$ and constant $k>0$, $\left\lvert\frac{z-z_1}{z-z_2}\right\rvert=k$ is the perpendicular bisector of $z_1z_2$ when $k=1$, and a circle for every $k\neq1$.

Why it is a circle. Put $\lvert z-z_1\rvert^2=k^2\lvert z-z_2\rvert^2$ and expand. The $z\bar z$ terms carry coefficients $1$ and $k^2$, so provided $k\neq1$ they do not cancel and you are left with $z\bar z+\bar az+a\bar z+b=0$ — the general circle. When $k=1$ they cancel exactly, the $z\bar z$ term vanishes, and a line is what remains.

Example 11 (JEE Main 2026, 24 Jan Shift 1). Let $S=\left\{z\in\mathbb C:\left\lvert\dfrac{z-6i}{z-2i}\right\rvert=1 \text{ and } \left\lvert\dfrac{z-8+2i}{z-4}\right\rvert=1\right\}$. Describe $S$.

▸ SHOW SOLUTION

Step 1: Both ratios equal $1$, so both conditions are perpendicular bisectors — no circles here at all.

Step 2: First condition. $\lvert z-6i\rvert=\lvert z-2i\rvert$: equidistant from $(0,6)$ and $(0,2)$, so $z$ lies on the horizontal line $y=4$.

Step 3: Second condition. $\lvert z-(8-2i)\rvert=\lvert z-4\rvert$: equidistant from $(8,-2)$ and $(4,0)$. Squaring,

$(x-8)^2+(y+2)^2=(x-4)^2+y^2\ \Longrightarrow\ -16x+64+4y+4=-8x+16$ $\Longrightarrow\ -8x+4y+52=0\ \Longrightarrow\ 2x-y-13=0.$

Step 4: Intersect. Substituting $y=4$ gives $2x=17$, so $x=\dfrac{17}{2}$.

Answer: $S$ is the single point $\boxed{\dfrac{17}{2}+4i}$ — two lines meet once.

Loci by angle

The constant-angle locus. For fixed $z_1\neq z_2$, $\arg\left(\frac{z-z_1}{z-z_2}\right)=\alpha$ is the set of points from which the segment $z_1z_2$ subtends the fixed angle $\alpha$.

$\alpha$	Locus
$0<\alpha<\pi$, $\alpha\neq\dfrac\pi2$	an arc of a circle through $z_1$ and $z_2$
$\alpha=\dfrac\pi2$	the circle with $z_1z_2$ as diameter
$\alpha=\pi$	the line segment between $z_1$ and $z_2$
$\alpha=0$	the line through $z_1,z_2$ outside the segment

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The radius of the arc follows from the sine rule: if the chord $z_1z_2$ has length $c$ and subtends $\alpha$, then $R=\dfrac{c}{2\sin\alpha}$.

Example 12. Let $z_1=10+6i$ and $z_2=4+6i$. If $z$ satisfies $\arg\left(\dfrac{z-z_1}{z-z_2}\right)=\dfrac{\pi}{4}$, show the locus is an arc and find its radius.

▸ SHOW SOLUTION

Step 1: The chord. $\lvert z_1-z_2\rvert=\lvert6\rvert=6$.

Step 2: Apply the sine rule. The chord subtends $\dfrac\pi4$ at every point of the arc, so

$R=\frac{c}{2\sin\alpha}=\frac{6}{2\sin\frac\pi4}=\frac{6}{2\cdot\frac{1}{\sqrt2}}=\boxed{3\sqrt2}$

Step 3: Sanity check. $3\sqrt2\approx4.24$, comfortably more than half the chord ($3$) — as any circle through both endpoints must be.

Example 13 (JEE Main 2026, 4 Apr Shift 1). Let $z$ satisfy $\lvert z+2\rvert=\lvert z-2\rvert$ and $\arg\left(\dfrac{z+3}{z-i}\right)=\dfrac{\pi}{4}$. Find $\lvert z\rvert^2$.

▸ SHOW SOLUTION

Step 1: Use the first condition. Equal distances from $-2$ and $2$ is the perpendicular bisector of that segment — the imaginary axis. So write $z=iy$ with $y$ real.

Step 2: Simplify the quotient.

$\frac{z+3}{z-i}=\frac{3+iy}{i(y-1)}.$

Multiply numerator and denominator by $-i$ to clear the imaginary unit below:

$=\frac{-i(3+iy)}{y-1}=\frac{y-3i}{y-1}=\frac{y}{y-1}+\left(\frac{-3}{y-1}\right)i.$

Step 3: Impose the argument. An argument of $\dfrac\pi4$ means the real and imaginary parts are equal:

$\frac{y}{y-1}=\frac{-3}{y-1}\ \Longrightarrow\ y=-3.$

Step 4: Check the quadrant. At $y=-3$ both parts equal $\dfrac{-3}{-4}=\dfrac34>0$, so the point is in the first quadrant and the argument really is $+\dfrac\pi4$, not $-\dfrac{3\pi}{4}$.

Step 5: $z=-3i$, so $\lvert z\rvert^2=\boxed{9}$

The quadrant check in Step 4 is not optional: $\dfrac{\text{Im}}{\text{Re}}=1$ is satisfied by both $\dfrac\pi4$ and $-\dfrac{3\pi}{4}$.

When a quotient is real or purely imaginary. For any $w\neq0$: $w$ is real $\iff w=\bar w \iff \operatorname{Im}w=0$, and $w$ is purely imaginary $\iff w=-\bar w \iff \operatorname{Re}w=0$. Applied to a quotient this is nearly always faster than separating real and imaginary parts by brute force: cross-multiply $\dfrac{A}{B}=\overline{\left(\dfrac{A}{B}\right)}$ into $A\bar B=\bar A B$, i.e. $\operatorname{Im}\left(A\bar B\right)=0$.

Example 14 (JEE Advanced 2022). Let $z$ be a non-real complex number for which

$\frac{2+3z+4z^{2}}{2-3z+4z^{2}}$

is real. Find $\lvert z\rvert^{2}$.

▸ SHOW SOLUTION

Step 1: Give the common part a name. Write $A=2+4z^{2}$, so the quotient is $\dfrac{A+3z}{A-3z}$.

Step 2: Impose reality by cross-multiplying. A number equals its own conjugate exactly when it is real, so

$(A+3z)\overline{(A-3z)}=\overline{(A+3z)}\,(A-3z).$

Step 3: Expand and cancel. The $A\bar A$ and $9z\bar z$ terms appear on both sides and cancel, leaving

$-3A\bar z+3\bar A z=-3\bar A z+3A\bar z \ \Longrightarrow\ 6\left(\bar A z-A\bar z\right)=0 \ \Longrightarrow\ \operatorname{Im}\left(\bar A z\right)=0.$

Step 4: Substitute $\bar A=2+4\bar z^{2}$.

$\bar A z=2z+4\bar z^{2}z=2z+4\bar z\left(\bar z z\right)=2z+4\lvert z\rvert^{2}\bar z,$

using $\bar z z=\lvert z\rvert^{2}$.

Step 5: Take the imaginary part. With $z=x+iy$ and $r=\lvert z\rvert^{2}$,

$\bar A z=\left(2x+4rx\right)+i\left(2y-4ry\right),\qquad \operatorname{Im}=y\left(2-4r\right).$

Step 6: Use "non-real". Non-real means $y\neq0$, so the bracket must vanish:

$2-4r=0\ \Longrightarrow\ \boxed{\lvert z\rvert^{2}=\frac12=0.50}$

Check. At $z=\dfrac{1+i}{2}$ we have $\lvert z\rvert^2=\dfrac12$, and the quotient evaluates to $7$ — real, as required. At $z=\dfrac{i}{\sqrt2}$ it evaluates to $-1$.

Ellipse, hyperbola and their degenerate cases

Let $c=\lvert z_1-z_2\rvert$ be the distance between two fixed points.

Condition	Locus
$\lvert z-z_1\rvert+\lvert z-z_2\rvert=k>c$	ellipse, foci $z_1,z_2$
$\lvert z-z_1\rvert+\lvert z-z_2\rvert=c$	the segment $z_1z_2$ (degenerate)
$\lvert z-z_1\rvert+\lvert z-z_2\rvert<c$	empty
$\bigl\lvert\,\lvert z-z_1\rvert-\lvert z-z_2\rvert\,\bigr\rvert=k<c$	hyperbola
$\bigl\lvert\,\lvert z-z_1\rvert-\lvert z-z_2\rvert\,\bigr\rvert=c$	the two rays outside the segment

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Example 15. If $\lvert z-a^2\rvert+\lvert z-2a\rvert=3$ always represents an ellipse, find the range of $a$ for $a>0$.

▸ SHOW SOLUTION

Step 1: The condition for a genuine ellipse is that the constant exceeds the focal distance:

$\lvert a^2-2a\rvert<3.$

Step 2: Split the modulus. $-3<a^2-2a<3$.

• Right: $a^2-2a-3<0\Rightarrow(a-3)(a+1)<0\Rightarrow-1<a<3$.
• Left: $a^2-2a+3>0$, whose discriminant is $4-12<0$, so it holds for every real $a$.

Step 3: Impose $a>0$.

$\boxed{a\in(0,3)}$

Example 16. Find the greatest and least values of $\lvert z-4\rvert$ given $\lvert z-1\rvert+\lvert z+3\rvert\le8$.

▸ SHOW SOLUTION

Step 1: Identify the region. Foci $1$ and $-3$, focal distance $4$, and $8>4$, so the region is the closed interior of an ellipse with $2a=8$.

Step 2: Extract the ellipse's data. Centre is the midpoint $-1$; $a=4$; $c=2$; so $b^2=a^2-c^2=12$. In coordinates,

$\frac{(x+1)^2}{16}+\frac{y^2}{12}\le1.$

Step 3: Maximise and minimise the distance to $(4,0)$. The point $(4,0)$ lies on the major axis, outside the ellipse. On the major axis the ellipse runs from $x=-5$ to $x=3$, so

$\lvert z-4\rvert_{\min}=\lvert3-4\rvert=1,\qquad \lvert z-4\rvert_{\max}=\lvert-5-4\rvert=9.$

Step 4: Confirm no interior critical point competes. Writing $d^2=(x-4)^2+12\left(1-\frac{(x+1)^2}{16}\right)$ and differentiating gives a stationary point at $x=19$, far outside $[-5,3]$, so the extremes really are at the vertices.

Answer: $\boxed{\lvert z-4\rvert\in[1,9]}$

Example 17. Identify the locus of $z$ satisfying $\bigl\lvert\,\lvert z+2i\rvert-\lvert z-2i\rvert\,\bigr\rvert=2$.

▸ SHOW SOLUTION

The two fixed points are $-2i$ and $2i$, a distance $c=4$ apart. The difference of the distances is constant and equals $2$, which is strictly less than $4$. That is precisely the definition of a $\boxed{\text{hyperbola}}$ with those two points as foci.

Had the constant been $4$ it would degenerate into two rays; had it exceeded $4$ the locus would be empty, since the triangle inequality forbids it.

Maximum and minimum modulus

Triangle inequality. For any $z_1,z_2$: $\bigl\lvert\,\lvert z_1\rvert-\lvert z_2\rvert\,\bigr\rvert\ \le\ \lvert z_1+z_2\rvert\ \le\ \lvert z_1\rvert+\lvert z_2\rvert.$ Equality on the right needs $z_1,z_2$ to point the same way; on the left, opposite ways.

Extremes on a circle. If $\lvert z-z_0\rvert=r$ then $\lvert z\rvert_{\min}=\bigl\lvert\,\lvert z_0\rvert-r\,\bigr\rvert,\qquad \lvert z\rvert_{\max}=\lvert z_0\rvert+r,$ and both are attained on the line joining $O$ to $z_0$.

Example 18. If $\lvert z-4+3i\rvert\le2$, find the least and greatest values of $\lvert z\rvert$.

▸ SHOW SOLUTION

Step 1: Read the region. A closed disc, centre $z_0=4-3i$, radius $2$.

Step 2: Distance from the origin to the centre. $\lvert z_0\rvert=\lvert4-3i\rvert=5$.

Step 3: The origin is outside (since $5>2$), so

$\lvert z\rvert_{\min}=5-2=3,\qquad \lvert z\rvert_{\max}=5+2=7.$

Answer: $\boxed{3\le\lvert z\rvert\le7}$

Example 19. If $z_1=24+7i$ and $\lvert z_2\rvert=6$, find the greatest and least values of $\lvert z_1+z_2\rvert$.

▸ SHOW SOLUTION

$\lvert z_1\rvert=\sqrt{576+49}=25$. By the triangle inequality,

$\lvert z_1\rvert-\lvert z_2\rvert\le\lvert z_1+z_2\rvert\le\lvert z_1\rvert+\lvert z_2\rvert,$

so $\boxed{19\le\lvert z_1+z_2\rvert\le31}$. Geometrically $z_1+z_2$ traces the circle of radius $6$ centred at $z_1$, and the extremes are its nearest and farthest points from the origin.

Example 20. Find the maximum value of $\lvert z\rvert$ when $\left\lvert z+\dfrac2z\right\rvert=2$.

▸ SHOW SOLUTION

Step 1: Bound below using the triangle inequality. Writing $r=\lvert z\rvert$,

$2=\left\lvert z+\frac2z\right\rvert\ \ge\ \bigl\lvert\,\lvert z\rvert-\tfrac{2}{\lvert z\rvert}\,\bigr\rvert=\left\lvert r-\frac2r\right\rvert.$

Step 2: Remove the modulus. $-2\le r-\dfrac2r\le2$, and since we want the largest $r$ we take the right-hand half:

$r-\frac2r\le2\ \Longrightarrow\ r^2-2r-2\le0.$

Step 3: Solve. The roots are $r=1\pm\sqrt3$, and $r>0$, so $0<r\le1+\sqrt3$.

Answer: $\lvert z\rvert_{\max}=\boxed{1+\sqrt3}$

Example 21 (JEE Main 2026, 2 Apr Shift 2 — the same idea). Let $z_1$ satisfy $\lvert z_1\rvert=12$ and $z_2$ satisfy $\lvert z_2-3-4i\rvert=5$. Find $\min\lvert z_1-z_2\rvert$.

▸ SHOW SOLUTION

Step 1: Two circles. $z_1$ runs on the circle centre $O$, radius $12$. $z_2$ runs on the circle centre $3+4i$, radius $5$.

Step 2: Note the second circle passes through the origin, because $\lvert3+4i\rvert=5$ equals its radius. So $\lvert z_2\rvert$ ranges over $[0,10]$.

Step 3: Minimise the gap. $\lvert z_1-z_2\rvert\ge\lvert z_1\rvert-\lvert z_2\rvert\ge12-10=2$, attained when $z_1$ and $z_2$ are both on the ray through $3+4i$.

Answer: $\boxed{2}$

A region cut by a half-plane. A condition of the form $z\bar a+\bar z a\le k$ (with $a$ complex, $k$ real) is always a half-plane, because $z\bar a+\bar z a=2\operatorname{Re}(z\bar a)$ is a real linear function of $x$ and $y$. Intersecting it with a disc leaves a circular segment, and the extremes of a distance over that segment sit either on the arc or on the straight edge — you must test both.

Example 22 (JEE Main 2024, 1 Feb Shift 1). Let $P=\{z:\lvert z+2-3i\rvert\le1\}$ and $Q=\{z:z(1+i)+\bar z(1-i)\le-8\}$. In $P\cap Q$, let $\lvert z-3+2i\rvert$ be greatest at $z_1$ and least at $z_2$. If $\lvert z_1\rvert^2+2\lvert z_2\rvert^2=\alpha+\beta\sqrt2$ with $\alpha,\beta$ integers, find $\alpha+\beta$.

▸ SHOW SOLUTION

Step 1: Read $P$. A closed disc, centre $-2+3i$, radius $1$.

Step 2: Read $Q$. With $z=x+iy$, $z(1+i)+\bar z(1-i)=2\operatorname{Re}\big(z(1+i)\big)=2(x-y)$, so $Q$ is the half-plane

$x-y\le-4.$

Step 3: Do they overlap? At the centre $(-2,3)$ we get $x-y=-5\le-4$, so the centre is inside $Q$. The distance from the centre to the boundary line $x-y+4=0$ is $\dfrac{\lvert-2-3+4\rvert}{\sqrt2}=\dfrac{1}{\sqrt2}<1$, so the line genuinely cuts the disc: $P\cap Q$ is a circular segment.

Step 4: The target point. $\lvert z-3+2i\rvert$ is the distance to $T=(3,-2)$, and

$\lvert T-\text{centre}\rvert=\sqrt{5^2+5^2}=5\sqrt2.$

Step 5: The maximum. The farthest point of the whole disc from $T$ is the centre pushed one unit away from $T$:

$z_1=\left(-2-\tfrac{1}{\sqrt2}\right)+\left(3+\tfrac{1}{\sqrt2}\right)i.$

Check it lies in $Q$: $x-y=-5-\sqrt2\approx-6.41\le-4$. It does, so the maximum is unaffected by the cut.

Step 6: The minimum — this is where the cut bites. The nearest point of the disc alone would be the centre pushed one unit towards $T$, at $(-1.29,2.29)$, where $x-y\approx-3.59>-4$ — outside $Q$. So the minimum must lie on the straight edge. Dropping a perpendicular from $T=(3,-2)$ to $x-y+4=0$ gives the foot

$z_2=-\tfrac32+\tfrac52 i,$

which is $\dfrac{1}{\sqrt2}<1$ from the centre, so it really is on the segment.

Step 7: Compute.

$\lvert z_1\rvert^2=\left(2+\tfrac{1}{\sqrt2}\right)^2+\left(3+\tfrac{1}{\sqrt2}\right)^2=14+5\sqrt2,\qquad \lvert z_2\rvert^2=\tfrac94+\tfrac{25}{4}=\tfrac{17}{2}.$

$\lvert z_1\rvert^2+2\lvert z_2\rvert^2=14+5\sqrt2+17=31+5\sqrt2.$

Step 8: $\alpha=31$, $\beta=5$, so $\alpha+\beta=\boxed{36}$

The lesson. The maximum was decided by the circle and the minimum by the line. Testing only the circle gives the minimum wrong; testing only the line gives the maximum wrong. Always check whether the unconstrained extreme survives the cut.

The shape JEE Main 2026 kept setting

Three of the 2026 questions were built the same way: one circle, one sum-of-distances condition, count the common points. The trick is always to check whether the sum-of-distances condition is a genuine ellipse or has degenerated.

Example 23 (JEE Main 2026, 8 Apr Shift 2). Find the number of $z\in\mathbb C$ satisfying

$\lvert z-(4+8i)\rvert=\sqrt{10}\quad\text{and}\quad \lvert z-(3+5i)\rvert+\lvert z-(5+11i)\rvert=4\sqrt5.$

▸ SHOW SOLUTION

Step 1: Identify the ellipse. The foci are $3+5i$ and $5+11i$. Their separation is

$\lvert(5+11i)-(3+5i)\rvert=\lvert2+6i\rvert=\sqrt{40}=2\sqrt{10},$

so $2c=2\sqrt{10}$, giving $c=\sqrt{10}$. The constant is $2a=4\sqrt5$, so $a=2\sqrt5$. Since $a=2\sqrt5\approx4.47$ exceeds $c\approx3.16$, this is a genuine ellipse.

Step 2: Find the semi-minor axis.

$b^2=a^2-c^2=20-10=10,\qquad b=\sqrt{10}.$

Step 3: Locate the circle. Its centre $4+8i$ is the midpoint of the foci, which is exactly the centre of the ellipse. So the two curves are concentric. Its radius is $\sqrt{10}$.

Step 4: Compare radius with semi-axes. The circle's radius is $\sqrt{10}$, and the semi-minor axis is also $\sqrt{10}$, while the semi-major axis is $2\sqrt5>\sqrt{10}$. So the circle is inscribed in the ellipse touching it exactly where the ellipse is nearest its centre — the two ends of the minor axis.

Answer: $\boxed{2}$ common points.

Why the numbers were chosen. Set $r$ against $b$: if $r<b$ the circle is strictly inside and there are $0$ solutions; if $r=b$ it touches at $2$ points; if $b<r<a$ it cuts at $4$; if $r=a$ it touches at $2$ again (the major-axis ends); if $r>a$ there are $0$. The examiner picked the one value of $r$ that gives exactly $2$ by tangency rather than by crossing.

Example 24 (JEE Main 2026, 28 Jan Shift 2). Let $A=\{z:\lvert z-2\rvert\le4\}$ and $B=\{z:\lvert z-2\rvert+\lvert z+2\rvert=8\}$. Describe $A\cap B$.

▸ SHOW SOLUTION

Step 1: $A$ is the closed disc of centre $2$, radius $4$.

Step 2: $B$. Foci $2$ and $-2$ are $4$ apart, and $8>4$, so $B$ is a genuine ellipse: centre $0$, $2a=8$ so $a=4$, $c=2$, $b^2=16-4=12$.

Step 3: Picture them together. The ellipse runs from $x=-4$ to $x=4$; the disc is centred at $x=2$ with radius $4$, so it spans $x\in[-2,6]$. The two overlap over the right-hand portion of the ellipse.

Step 4: The intersection is therefore an arc of the ellipse, not a finite set of points — the part of $B$ lying within distance $4$ of the point $2$.

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