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JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Complex Numbers

Let

i = - 1

. If

\frac{( - 1 + i 3 ) ^{21}}{( 1 - i ) ^{24}} + \frac{( 1 + i 3 ) ^{21}}{( 1 + i ) ^{24}} = k

, and

n = [∣ k ∣]

be the greatest integral part of

∣ k ∣

. Then

\sum_{j = 0}^{n + 5} (j + 5)^{2} - \sum_{j = 0}^{n + 5} (j + 5)

is equal to

Answer: 310

Solution

In polar form

- 1 + i 3 = 2 e^{i 2 π /3}

and

1 - i = 2 e^{- iπ /4}

, so

\frac{( - 1 + i 3 ) ^{21}}{( 1 - i ) ^{24}} = \frac{2 ^{21} e ^{i 14 π}}{2 ^{12} e ^{- i 6 π}} = 2^{9} = 512

. Likewise

1 + i 3 = 2 e^{iπ /3}

and

1 + i = 2 e^{iπ /4}

give

\frac{( 1 + i 3 ) ^{21}}{( 1 + i ) ^{24}} = \frac{2 ^{21} e ^{i 7 π}}{2 ^{12} e ^{i 6 π}} = 2^{9} e^{iπ} = - 512

. Hence

k = 512 - 512 = 0

and

n = 0

, so the required difference is

j = 0 \sum 5 [(j + 5)^{2} - (j + 5)] = j = 0 \sum 5 (j + 5) (j + 4) = 20 + 30 + 42 + 56 + 72 + 90,

which equals

310

.

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer

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