Ano ang ibig sabihin ng (3 + i) ^ (1/3) sa isang + bi form?

Sagot:

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

Paliwanag:

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

Kaya

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

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

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

Mula noon # 3 + i # nasa Q1, ang punong-guro ng kubo na ito ng # 3 + i # ay nasa Q1 rin.

Ang dalawang iba pang mga cube roots ng # 3 + i # ay ipinahayag gamit ang primitive Complex cube root ng pagkakaisa #omega = -1 / 2 + sqrt (3) / 2 i #:

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

(6) (10) cos (1/3 arctan (1/3) + (2pi) / 3) + root (6) (10) kasalanan (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) cos (1/3 arctan (1/3) + (4pi) / 3) + root (6) (10) kasalanan (1/3 arctan (1/3) + (4pi) / 3) i #