JEE Mains Maths · Complex Numbers

Polar Form, Argument and De Moivre

Writing a complex number by its length and angle, so that products add angles, powers multiply them, and multiplying by a unit complex number rotates the plane.

All Complex Numbers notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs, nineteen of them multiple choice. Eight find a modulus or a principal argument, eight raise a number to a high power, and eight use rotation to settle a triangle or a square. Each one is quicker in polar form than in x and y. Three ideas cover the page.

Concept 1 of 3: Polar form and the principal argument

Any

z\neq0

r(\cos\theta+i\sin\theta)

, written

re^{i\theta}

, with

r=|z|

and

\theta=\arg z

. The principal argument lies in

(-\pi,\pi]

: find the reference angle

\alpha=\tan^{-1}\left|\frac yx\right|

, then place it by the quadrant of

z

. Multiplying multiplies moduli and adds arguments, dividing subtracts them, and conjugating negates the argument.

Definition

$z=re^{i\theta}$ with $r=|z|$ , $\theta=\arg z\in(-\pi,\pi]$ .
By quadrant: $\alpha$ , $\pi-\alpha$ , $-\pi+\alpha$ , $-\alpha$ .
$\arg(z_1z_2)=\arg z_1+\arg z_2$ and $\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2$ , up to a multiple of $2\pi$ .
$\arg\bar z=-\arg z$ ; for $|z|=1$ , $\frac1z=\bar z$ .

Polar form

z=r(\cos\theta+i\sin\theta)=re^{i\theta}

Worked example

Find the modulus and principal argument of

-1-\sqrt3i

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q156Moderate

Example 1 · Complex Numbers · Polar Form, Argument and De Moivre

Let

r

and

\theta

respectively be the modulus and amplitude of the complex number

z = 2 - i\left( 2\tan\frac{5\pi}{8} \right)

, then

(r,\theta)

is equal to

The inverse tangent is not the argument

\tan^{-1}\frac yx

lands in

\left(-\frac\pi2,\frac\pi2\right)

. For a point with

x<0

, add or subtract

\pi

to reach the right quadrant.

Drill 7 more on polar form and the principal argument

Concept 2 of 3: Powers by De Moivre's theorem

Raising

re^{i\theta}

to the power

n

raises the modulus to

r^n

and multiplies the angle by

n

. So write the base in polar form, multiply the angle, then take away multiples of

2\pi

. A sum like

z^n+\bar z^n

2\,\mathrm{Re}(z^n)=2r^n\cos n\theta

, and

w^n

is real exactly when

n\theta

is a multiple of

\pi

Definition

$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$ .
$1+i=\sqrt2e^{i\pi/4}$ , $1+\sqrt3i=2e^{i\pi/3}$ , $\sqrt3+i=2e^{i\pi/6}$ .
$z^n+\bar z^n=2r^n\cos n\theta$ .
$1+\cos\theta+i\sin\theta=2\cos\frac\theta2\,e^{i\theta/2}$ .

De Moivre's theorem

(re^{i\theta})^n=r^ne^{in\theta}

Worked example

Find

(1+i)^{10}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q67Moderate

Example 2 · Complex Numbers · Polar Form, Argument and De Moivre

z =\frac{\sqrt{3}}{2}+\frac{i}{2},i =\sqrt{- 1}

, then

\left( z^{201}- i \right)^{8}

is equal to

Reduce the angle, and raise the modulus

\frac{201\pi}6

is not a principal argument: take away

32\pi

first. And a base of modulus 2 raised to the 21st power carries a factor

2^{21}

that is easy to drop.

Drill 7 more on powers by de moivre's theorem

Concept 3 of 3: Rotation and triangles

Multiplying by

e^{i\alpha}

turns the plane about the origin through

\alpha

i

is a quarter turn anticlockwise,

-i

a quarter turn clockwise. To turn about a point

a

, shift first:

w-a=(z-a)e^{i\alpha}

. In a triangle,

\frac{z_3-z_1}{z_2-z_1}

has modulus equal to the ratio of the two sides at

z_1

and argument equal to the angle between them.

Definition

About the origin: $w=ze^{i\alpha}$ ; a quarter turn is $iz$ .
About $a$ : $w=a+(z-a)e^{i\alpha}$ .
$\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|}{|z_2-z_1|}e^{i\theta}$ , $\theta$ the angle at $z_1$ .
Area of the triangle $0,z_1,z_2$ : $\frac12\left|\mathrm{Im}(\bar z_1z_2)\right|$ .

Angle at a vertex

\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|}{|z_2-z_1|}\,e^{i\theta}

Worked example

A square has adjacent vertices

0

and

3+i

, taken anticlockwise. Find the other two.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q68Moderate

Example 3 · Complex Numbers · Polar Form, Argument and De Moivre

Let

w_{1}

be the point obtained by the rotation of

z_{1}= 5 + 4i

about the origin through a right angle in the anticlockwise direction, and

w_{2}

be the point obtained by the rotation of

z_{2}= 3 + 5i

about the origin through a right angle in the clockwise direction. Then the principal argument of

w_{1}-w_{2}

is equal to

Turn about the right point

ze^{i\alpha}

turns about the origin only. For a turn about

a

, subtract

a

, rotate, then add

a

back; clockwise uses

e^{-i\alpha}

Drill 7 more on rotation and triangles

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 (3)

Polar form and the principal argument
Polar form
$z=r(\cos\theta+i\sin\theta)=re^{i\theta}$
Powers by De Moivre's theorem
De Moivre's theorem
$(re^{i\theta})^n=r^ne^{in\theta}$
Rotation and triangles
Angle at a vertex
$\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|}{|z_2-z_1|}\,e^{i\theta}$

Watch out for (3)

The inverse tangent is not the argument
→ Polar form and the principal argument
Reduce the angle, and raise the modulus
→ Powers by De Moivre's theorem
Turn about the right point
→ Rotation and triangles

Test yourself on Complex Numbers

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter test Drill all 24 questions