Apakah (3 + i) ^ (1/3) sama dengan bentuk + bi?

Jawapan:

#root (6) (10) cos (1/3 arctan (1/3)) + root (6) (10) sin (1/3 arctan (1/3)

Penjelasan:

# 3 + i = sqrt (10) (cos (alpha) + i sin (alpha)) # di mana #alpha = arctan (1/3) #

Jadi

#root (3) (3 + i) = root (3) (sqrt (10)) (cos (alpha / 3) + i sin (alpha / 3)

# = root (6) (10) (cos (1/3 arctan (1/3)) + i sin (1/3 arctan (1/3)) #

# = root (6) (10) cos (1/3 arctan (1/3)) + root (6) (10) sin (1/3 arctan (1/3)

Sejak # 3 + i # berada pada Q1, akar kiub utama ini # 3 + i # juga dalam Q1.

Kedua-dua akar kiub lain dari # 3 + i # boleh diungkapkan menggunakan akar kubus primitif Kompleks perpaduan #omega = -1 / 2 + sqrt (3) / 2 i #:

#omega (root (6) (10) cos (1/3 arctan (1/3)) + root (6) (10) sin (1/3 arctan (1/3)

(6) (10) sin (1/3 arctan (1/3) + (2pi) / 3) i #

# omega ^ 2 (root (6) (10) cos (1/3 arctan (1/3)) + root (6) (10) sin (1/3 arctan (1/3)

(6) (10) sin (1/3 arctan (1/3) + (4pi) / 3) i #