Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If (1+i root 3)^9 = a+ib, then b is equal to (1) 1 (2) 256 (3) 0 (4) 9^3

Solution:

(1+i√3)9 = a+ib

= 29 [(½)+(√3/2)i ]9

= 29 [ cos π/3 + i sin π/3]9

Apply de moivre’s theorem

= 29 [ cos 9π/3 + i sin 9π/3]

= 29 [ cos 9π/3 + i sin 9π/3]

= 29 [ cos 3π+ i sin 3π]

= 29 [ cos π+ i sin π]

= 29 (-1+0)

= -29

Imaginary part is 0.

Hence option (3) is the answer.

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