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Question : 71 of 160
Marks:
+1 ,
-0
Solution:
( √ 3 + i ) 7 + ( √ 3 − i ) 7 Let
‌ ‌ √ 3 + i = z z = ( √ 3 + i ) = r ( cos ‌ θ + i s i n ‌ θ ) Then,
‌ ‌ r ‌ cos ‌ θ = √ 3 . . . (i)
r s i n ‌ θ = 1 . . . (ii)
Eq. (i) + Eq. (ii), we get
‌ r 2 ( s i n ‌ 2 θ + cos 2 θ ) = 3 + 1 ⇒ r 2 = 4 ⇒ r = 2 Eq. (ii)
÷ Eq. (i), we get
‌ ⇒ ‌ ‌ tan ‌ θ = ‌ = tan ‌ ‌ ⇒ ‌ ‌ θ = ‌ ‌ ‌ Then, ‌ z = ( √ 3 + i ) = 2 ( cos ‌ + i s i n ‌ ‌ ) ‌ ⇒ ‌ ‌ z = ( √ 3 − i ) = 2 ( cos ‌ − i s i n ‌ ‌ ) ‌ ‌ Then, ‌ ( √ 3 + i ) 7 + ( √ 3 − i ) 7 ‌ = [ 2 { cos ‌ + i s i n ‌ ‌ } ] 7 ‌ ‌ ‌ + [ 2 { cos ‌ − i s i n ‌ ‌ } ] 7 ‌ = 2 7 { cos ‌ + i s i n ‌ ‌ } + 2 7 ‌ ‌ ‌ { cos ‌ − i s i n ‌ ‌ } (By De-moiver theorem)
‌ = 2 7 { cos ‌ + i s i n ‌ ‌ + cos ‌ − i s i n ‌ ‌ } ‌ = 2 7 ⋅ 2 ‌ cos ‌ ‌ = 2 8 ⋅ cos ( π + π ∕ 6 ) ‌ = − 2 8 ⋅ cos ‌ ‌ = − 2 8 ⋅ ‌ ‌ = − 2 7 ⋅ √ 3 = − 128 √ 3
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