Is De Moivre's Theorem part of the EAPCET Mathematics syllabus?

Yes. De Moivre's Theorem is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Mathematics questions are there in EAPCET?

80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.

What mistakes do students make in De Moivre's Theorem?

Raising only the cos θ + i sin θ part to the power and forgetting to raise the modulus r to the power n as well. Finding a single nth root of a number; add 2kπ to the argument and take k = 0, 1, …, n − 1 to get all n roots. Reducing ωⁿ with the wrong remainder; for example ω¹⁰⁰ = ω because 100 leaves remainder 1 on division by 3. Using 1 + ω + ω² = 0 for ω = 1; it holds only for the non-real cube roots of unity.

EAPCET Mathematics · Intermediate second year · Paper IIA

De Moivre's Theorem

De Moivre's theorem raises a complex number in polar form to a power, and gives every nth root of a complex number, including the roots of unity.

Subject: Mathematics
Taught in: Intermediate second year · Paper IIA
Mathematics in the paper: 80 of 160 questions

Key formulas and results

$(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta$
Cube roots of unity are $1, \omega, \omega^{2}$ , with $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$
$\omega = \dfrac{-1 + i\sqrt{3}}{2}$
The $n$ th roots of unity add up to 0 (for $n > 1$ )
The $n$ th roots of unity lie on the unit circle, at the vertices of a regular $n$ -gon
The $n$ th roots of unity: $\cos\dfrac{2k\pi}{n} + i\sin\dfrac{2k\pi}{n}$ , $k = 0, 1, \dots, n - 1$
$\dfrac{1}{\cos\theta + i\sin\theta} = \cos\theta - i\sin\theta$
$\omega^{2} = \dfrac{-1 - i\sqrt{3}}{2}$ , and $\omega^{3n} = 1$ , $\omega^{3n+1} = \omega$ , $\omega^{3n+2} = \omega^{2}$
If $1, \alpha_{1}, \dots, \alpha_{n-1}$ are the $n$ th roots of unity: $z^{n} - 1 = (z - 1)(z - \alpha_{1})\cdots(z - \alpha_{n-1})$

What this chapter covers

Statement and applications of De Moivre's theorem
nth roots of unity
Cube roots of unity
Roots of a complex number

How it is usually asked

Evaluate powers such as (1 + i)ⁿ or (√3 + i)ⁿ
Simplify expressions involving ω
Find the roots of zⁿ = a

Understanding De Moivre's Theorem

Any complex number can be written in polar form z = r(cos θ + i sin θ), where r is the modulus and θ the argument. De Moivre’s theorem says (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, so to raise z to a power you raise the modulus to that power and multiply the argument by it: zⁿ = rⁿ(cos nθ + i sin nθ). This replaces long binomial expansions with one line.

The theorem also works in reverse to find roots. The equation zⁿ = 1 has exactly n roots, cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, …, n − 1. They lie on the unit circle at the vertices of a regular n-gon, they add up to zero for n > 1, and their product is (−1)ⁿ⁻¹. The roots of zⁿ = a are found the same way after writing a in polar form.

The cube roots of unity 1, ω and ω² are the most tested case. Remember ω³ = 1 and 1 + ω + ω² = 0, and reduce any power of ω using the remainder of the exponent on division by 3. EAPCET questions evaluate powers like (1 + i)ⁿ or (√3 + i)ⁿ, simplify expressions in ω, or locate the nth roots of a number.

Worked example

EAPCET-style question

The value of $(\sqrt{3} + i)^{6}$ is

(1)64
(2)−64
(3)64i
(4)−64i

Solution

$\sqrt{3} + i$ has modulus $\sqrt{3 + 1} = 2$ and argument $\tfrac{\pi}{6}$ .
So $\sqrt{3} + i = 2\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)$ .
By De Moivre, $(\sqrt{3} + i)^{6} = 2^{6}(\cos\pi + i\sin\pi)$ .
$\cos\pi = -1$ and $\sin\pi = 0$ , so the value is $-64$ .

Answer: −64

Shortcuts and results to remember

De Moivre needs the exact form $\cos\theta + i\sin\theta$ : $(\sin\theta + i\cos\theta)^{n}$ is not $\sin n\theta + i\cos n\theta$ ; first rewrite it as $\cos\left(\tfrac{\pi}{2} - \theta\right) + i\sin\left(\tfrac{\pi}{2} - \theta\right)$ .
$(\cos\theta + i\sin\varphi)^{n}$ is not $\cos n\theta + i\sin n\varphi$ when $\theta \ne \varphi$ .
The roots of $x^{2} + x + 1 = 0$ are $\omega$ and $\omega^{2}$ ; the roots of $x^{2} - x + 1 = 0$ are $-\omega$ and $-\omega^{2}$ .

Common mistakes to avoid

Raising only the cos θ + i sin θ part to the power and forgetting to raise the modulus r to the power n as well.
Finding a single nth root of a number; add 2kπ to the argument and take k = 0, 1, …, n − 1 to get all n roots.
Reducing ωⁿ with the wrong remainder; for example ω¹⁰⁰ = ω because 100 leaves remainder 1 on division by 3.
Using 1 + ω + ω² = 0 for ω = 1; it holds only for the non-real cube roots of unity.

How to prepare it

Convert to polar form before raising to a power — expanding binomially is slower and riskier.

Once the formulas feel familiar, test them the way the exam does: practise De Moivre's Theorem questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise De Moivre's Theorem chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is De Moivre's Theorem part of the EAPCET Mathematics syllabus?: Yes. De Moivre's Theorem is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare De Moivre's Theorem for EAPCET?: Convert to polar form before raising to a power — expanding binomially is slower and riskier.
How many Mathematics questions are there in EAPCET?: 80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in De Moivre's Theorem?: Raising only the cos θ + i sin θ part to the power and forgetting to raise the modulus r to the power n as well. Finding a single nth root of a number; add 2kπ to the argument and take k = 0, 1, …, n − 1 to get all n roots. Reducing ωⁿ with the wrong remainder; for example ω¹⁰⁰ = ω because 100 leaves remainder 1 on division by 3. Using 1 + ω + ω² = 0 for ω = 1; it holds only for the non-real cube roots of unity.