If $\omega\ne1$ is a cube root of unity, show that (i) $(1-\omega+\omega^2)^6+(1+\omega-\omega^2)^6=128$. (ii) $(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)\cdots(1+\omega^{2^{11}})=1$.

If ω≠1 is a cube root of unity, show that (i) (1-ω+ω²)⁶+(1+ω-ω²)⁶=128. (ii) (1+ω)(1+ω²)(1+ω⁴)(1+ω⁸)…(1+ω^2¹¹)=

Exercise 2.8 · Q8

\omega\ne1

(i)

(1-\omega+\omega^2)^6+(1+\omega-\omega^2)^6=128

(ii)

(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)\cdots(1+\omega^{2^{11}})=1

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Concept understanding — Roots of Complex Numbers / nth Roots of Unity

$n$ th roots of a complex number. If $\omega=\rho(\cos\phi+i\sin\phi)$ satisfies $\omega^n=z$ where $z=r(\cos\theta+i\sin\theta)$ , write $z$ with its general argument $\theta+2k\pi$ (since a single value of $\theta$ would miss roots) and apply de Moivre's theorem to $\omega^n$ : comparing modulus and argument gives $\rho^n=r$ and $n\phi=\theta+2k\pi$ , so

\rho=r^{1/n},\qquad \phi=\frac{\theta+2k\pi}{n}.

Hence the $n$ th roots of $z=r(\cos\theta+i\sin\theta)$ are

z^{1/n}=r^{1/n}\left(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\right),\qquad k=0,1,2,\dots,n-1.

Although $k$ can be any integer, only $k=0,1,\dots,n-1$ give distinct values (larger $k$ repeats the same $n$ roots cyclically). Geometrically, all $n$ roots share the modulus $r^{1/n}$ , so they lie on a circle of radius $r^{1/n}$ centred at the origin, equally spaced at angular intervals of $\dfrac{2\pi}n$ — i.e. at the vertices of a regular $n$ -gon.

$n$ th roots of unity. Setting $z=1=\cos0+i\sin0$ specialises the formula to

z=\cos\frac{2k\pi}n+i\sin\frac{2k\pi}n=e^{2k\pi i/n},\qquad k=0,1,\dots,n-1.

Writing $\omega=e^{2\pi i/n}$ (the primitive $n$ th root), the $n$ roots are exactly $1,\omega,\omega^2,\dots,\omega^{n-1}$ — a geometric progression with common ratio $\omega$ — and they are the vertices of a regular $n$ -gon inscribed in the unit circle.

Standing facts about the $n$ th roots of unity (all provable from the GP sum/product formulas):

Sum $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$ (a finite GP with ratio $\omega\ne1$ , sum $\dfrac{\omega^n-1}{\omega-1}=\dfrac{1-1}{\omega-1}=0$ ).
Product $1\cdot\omega\cdot\omega^2\cdots\omega^{n-1}=(-1)^{n-1}$ .
They all satisfy $|z|=1$ and $z^n=1$ . …

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Part (i) reduces each bracket to a small multiple of a power of $\omega$ using $1+\omega+\omega^2=0$ , then uses $\omega^3=1$ to collapse the 6th powers. Part (ii) uses $\omega^3=1$ to see that the exponents $2^r$ cycle through the residues $1,2$ mod $3$ (since $2\equiv-1\pmod3$ ), pairing the factors into copies of $(1+\omega)(1+\omega^2)=1$ .

Step 1. (i) Simplify $1-\omega+\omega^2$ using $1+\omega+\omega^2=0\Rightarrow1+\omega^2=-\omega$ .

1-\omega+\omega^2=(1+\omega^2)-\omega=-\omega-\omega=-2\omega.

Step 2. (i) Simplify $1+\omega-\omega^2$ using $1+\omega=-\omega^2$ .

1+\omega-\omega^2=(1+\omega)-\omega^2=-\omega^2-\omega^2=-2\omega^2.

Step 3. (i) Raise both to the 6th power.

(-2\omega)^6=2^6\omega^6=64(\omega^3)^2=64(1)^2=64.

(-2\omega^2)^6=2^6\omega^{12}=64(\omega^3)^4=64(1)^4=64.

Step 4. (i) Add the two results.

(1-\omega+\omega^2)^6+(1+\omega-\omega^2)^6=64+64=128.

Step 5. (ii) Reduce every exponent $2^r$ modulo $3$ . Since $\omega^3=1$ , $\omega^{2^r}=\omega^{2^r\bmod3}$ . As $2\equiv-1\pmod3$ , $2^r\equiv(-1)^r\pmod3$ , so $2^r\bmod3$ alternates $2,1,2,1,\dots$ for $r=1,2,3,4,\dots$ (and $2^0=1\equiv1$ ). Listing the factors $1+\omega^{2^r}$ for $r=0,1,\dots,11$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)^{3/4}$
(a) $1$
(b) $-2$
(c) $2$
(d) $-1$
›Reveal solutionSolution
Lists the four values of the fractional power by de Moivre's theorem and multiplies them, using that the sum of their arguments is a multiple of $2\pi$ .

Write $z=\cos\dfrac\pi3+i\sin\dfrac\pi3=e^{i\pi/3}$ .

By the generalised de Moivre theorem, the four values of $z^{3/4}$ are $w_k=\cos\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)+i\sin\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)$ for $k=0,1,2,3$ , i.e. arguments $\theta_k=\dfrac{\pi+6k\pi}4$ .

$\theta_0=\dfrac\pi4$ , $\theta_1=\dfrac{7\pi}4$ , $\theta_2=\dfrac{13\pi}4$ , $\theta_3=\dfrac{19\pi}4$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The square root of i are :$
(a) $\pm\dfrac12(1+i)$
(b) $\pm\dfrac{1}{\sqrt2}(1+i)$
(c) $\pm\dfrac12(1-i)$
(d) $\pm\dfrac{1}{\sqrt2}(1-i)$
›Reveal solutionSolution
Writing the square root as $x+iy$ and equating real and imaginary parts of $(x+iy)^2=i$ pins down $x=y=1/\sqrt2$ .

Let $\sqrt i=x+iy$ for real $x,y$ . Then $(x+iy)^2=i$ .

Expand: $x^2-y^2+2xyi=0+1\cdot i$ .

Equate real parts: $x^2-y^2=0\Rightarrow x^2=y^2\Rightarrow x=\pm y$ .

Equate imaginary parts: $2xy=1$ , so $xy>0$ , meaning $x$ and $y$ have the same sign — hence $x=y$ (not $x=-y$ ). …
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CBSE 2018Set ANNUAL1 markMCQ
$\omega$
(a) $-16$
(b) $0$
(c) $-32$
(d) $32$
›Reveal solutionSolution
Rewriting each bracket using $1+\omega+\omega^2=0$ as $-2\omega$ and $-2\omega^2$ , then simplifying powers with $\omega^3=1$ , gives the value $-16$ .

Recall the key property of a cube root of unity $\omega\neq1$ : $1+\omega+\omega^2=0$ and $\omega^3=1$ .

Rewrite $1-\omega+\omega^2 = (1+\omega+\omega^2) - 2\omega = 0-2\omega = -2\omega$ .

Rewrite $1+\omega-\omega^2 = (1+\omega+\omega^2) - 2\omega^2 = 0-2\omega^2 = -2\omega^2$ .

So the expression becomes $(-2\omega)^4 + (-2\omega^2)^4 = 16\omega^4 + 16\omega^8$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\omega$
(a) 9
(b) $-9$
(c) 16
(d) 32
›Reveal solutionSolution
Reduce all powers of $\omega$ mod 3, use $(1-\omega)(1-\omega^2)=3$ (a standard cube-root-of-unity identity), then square it.

$\omega$ is a complex cube root of unity, so $\omega^3=1$ and $1+\omega+\omega^2=0$ .

Reduce the exponents modulo 3: $\omega^4=\omega^{3+1}=\omega$ , and $\omega^8=\omega^{6+2}=\omega^2$ .

So the product becomes $(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2) = \big[(1-\omega)(1-\omega^2)\big]^2$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$\left[\dfrac{-1+i\sqrt3}{2}\right]^{100} + \left[\dfrac{-1-i\sqrt3}{2}\right]^{100}$
(a) $2$
(b) $0$
(c) $-1$
(d) $1$
›Reveal solutionSolution
The expression equals $\omega+\omega^2=-1$ , where $\omega$ is a complex cube root of unity.

Let $\omega=\dfrac{-1+i\sqrt3}{2}$ ; then $\omega^2=\dfrac{-1-i\sqrt3}{2}$ , with $\omega^3=1$ and $1+\omega+\omega^2=0$ .

$\omega^{100}=\omega^{99}\cdot\omega=(\omega^3)^{33}\cdot\omega=1^{33}\cdot\omega=\omega$ .

$(\omega^2)^{100}=\omega^{200}=\omega^{198}\cdot\omega^2=(\omega^3)^{66}\cdot\omega^2=\omega^2$ . …
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Q.If $\omega\ne1$ is a cube root of unity, show that

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