In the multiplicative group of $n^{th}$ roots of unity, the inverse of $\omega^k$ is $(k < n)$ : (a) $\omega^{\frac{1}{k}}$ (b) $\omega^{-1}$ (c) $\omega^{n-k}$ (d) $\omega^{\frac{n}{k}}$

> [!TLDR] > In the group of $n^{th}$ roots of unity, $(\omega^k)^{-1}=\omega^{n-k}$. > > 1. The $n^{th}$ roots of unity form a cyclic group of order $n$ under multiplication, with identity element $1=\omega^0=\omega^n$. > 2. For an element $\omega^k$ ($k 3. Hence the inverse of $\omega^k$ is $\omega^{n-k}$; equivalently, since all $n^{th}$ roots of unity lie on the unit circle, $\omega^{n-k}=\overline{\omega^k}=(\omega^k)^{-1}$. …

In the multiplicative group of nth roots of unity, – Class 12 Mathematics

Question 52 of 84

n^{th}

(a)

\omega^{\frac{1}{k}}

(b)

\omega^{-1}

(c)

\omega^{n-k}

(d)

\omega^{\frac{n}{k}}

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Concept understanding — Modular Arithmetic

Modular Arithmetic — The Arithmetic of Remainders

You already know modular arithmetic. You just don't know you know it.

Think about a clock. When it's 10 AM and you add 5 hours, you get 3 PM — not 15 o'clock. The clock "wraps around" after 12. That wrapping is the entire idea of modular arithmetic. We care only about the remainder after division, not the full number.

The Intuition: "What's left over?"

Take any two whole numbers and divide one by the other. The result is a quotient (how many times it fits) and a remainder (what's left). Modular arithmetic is the study of that remainder.

For example: 17 divided by 5 gives quotient 3 and remainder 2. In modular language, we say:

17 \equiv 2 \pmod{5}

Read this as: "17 is congruent to 2 modulo 5." It means 17 and 2 leave the same remainder when divided by 5.

Note

The word "modulo" comes from Latin — meaning "with respect to the modulus." The modulus is the number you divide by (here, 5).

The Precise Definition

Let $a$ , $b$ , and $m$ be integers, with $m > 0$ . We say:

a \equiv b \pmod{m}

if and only if $m$ divides $(a - b)$ exactly — that is, $a - b = m \cdot k$ for some integer $k$ .

Equivalently: $a$ and $b$ have the same remainder when divided by $m$ .

a \equiv b \pmod{m} \iff m \mid (a - b)

Examples to Lock It In

Statement	Why it's true
$23 \equiv 3 \pmod{5}$	$23 - 3 = 20$ , and $5$ divides $20$
$14 \equiv 2 \pmod{12}$	$14 - 2 = 12$ , and $12$ divides $12$
$100 \equiv 1 \pmod{3}$	$100 - 1 = 99$ , and $3$ divides $99$
$7 \equiv 0 \pmod{7}$	$7 - 0 = 7$ , and $7$ divides $7$

Notice the last one: any number is congruent to 0 modulo itself. That's just saying the remainder when you divide a number by itself is 0.

What Modular Arithmetic Doesn't Care About

Two numbers that are congruent modulo $m$ are considered "the same" for all modular purposes. So:

... , -10, -5, 0, 5, 10, 15, ...

are all the same modulo 5. They form an equivalence class — a set of numbers that all share the same remainder.

Watch out

A common mistake: thinking $a \equiv b \pmod{m}$ means $a$ divided by $m$ gives remainder $b$ . That's only true if $0 \leq b < m$ . The definition is about the difference being divisible by $m$ , not about the remainder itself.

Why It Matters …

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In the group of $n^{th}$ roots of unity, $(\omega^k)^{-1}=\omega^{n-k}$ .

The $n^{th}$ roots of unity form a cyclic group of order $n$ under multiplication, with identity element $1=\omega^0=\omega^n$ .
For an element $\omega^k$ ( $k<n$ ), we need $m$ such that $\omega^k\cdot\omega^m=\omega^n=1$ , i.e. $k+m\equiv0\pmod n$ , so $m=n-k$ .
Hence the inverse of $\omega^k$ is $\omega^{n-k}$ ; equivalently, since all $n^{th}$ roots of unity lie on the unit circle, $\omega^{n-k}=\overline{\omega^k}=(\omega^k)^{-1}$ . …

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Q.In the multiplicative group of nthn^{th}nth roots of unity, the inverse of ωk\omega^kωk is (k<n)(k < n)(k<n) :

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Q.In the multiplicative group of $n^{th}$ roots of unity, the inverse of $\omega^k$ is $(k < n)$ :

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