Apa yang (3 + i) ^ (1/3) sama dengan dalam bentuk + bi?

Menjawab:

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

Penjelasan:

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

Begitu

#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)) i #

Sejak # 3 + i # di Q1, akar pangkat tiga utama dari # 3 + i # juga di Q1.

Dua akar pangkat tiga lainnya # 3 + i # dapat diekspresikan menggunakan akar kubus Kompleks primitif persatuan #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)) i) #

# = root (6) (10) cos (1/3 arctan (1/3) + (2pi) / 3) + root (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)) i) #

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