2 (cos 11 п/12+i sin 11 п/12) ^2

Пусть x = 11 п/12 для удобности написания

2 (cos x + i sin x) ^2 = 2 (cos^2 x + 2*i * (cos x) * (sin x) + sin^2 x)

cos^2 x + sin^2 x = 1 поэтому = > 2 (cos^2 x + 2*i * (cos x) * (sin x) + sin^2 x) =

=2 (1+2*i * (cos x) * (sin x))

2 * (sin x) * (cos x) = sin 2x

теперь подставим x

sin 2x = sin 22 п/12 = sin 11 п/6 = - 1/2

2 (1+2*i * (cos x) * (sin x)) = 2 (1+i*sin 2x) = 2 (1+i * (-1/2)) = 2-i

Ответ: 2-i