Let \(z=\frac{1-i \sqrt{3}}{2}, i=\sqrt{-1}\). Then the value of \(21+\left(z+\frac{1}{z}\right)^{3}+\left(z^{2}+\frac{1}{z^{2}}\right)^{3}+\left(z^{3}+\frac{1}{z^{3}}\right)^{3}+\ldots \ldots+\left(z^{21}+\frac{1}{z^{21}}\right)^{3}\) is ____.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Complex Numbers

Let

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

. Then the value of

21 + (z + \frac{1}{z})^{3} + (z^{2} + \frac{1}{z ^{2}})^{3} + (z^{3} + \frac{1}{z ^{3}})^{3} + \dots\dots + (z^{21} + \frac{1}{z ^{21}})^{3}

is ____.

Answer: 13

Here

∣ z ∣ = 1

and

ar g z = - \frac{π}{3}

, so

z = e^{- iπ /3}

and

z^{k} + \frac{1}{z ^{k}} = 2 cos \frac{k π}{3},

which repeats with period

6

k

. For

k = 1, 2, 3, 4, 5, 6

these values are

1, - 1, - 2, - 1, 1, 2

, so their cubes are

1, - 1, - 8, - 1, 1, 8

and add to

0

over any six consecutive values of

k

.
The terms

k = 1

k = 18

make up three such blocks and contribute

0

, while

k = 19, 20, 21

repeat the pattern of

k = 1, 2, 3

and contribute

1 - 1 - 8 = - 8

.
The value of the expression is

21 - 8 = 13

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer