以下の複素数でドモアブルの定理を使いz^413を計

質問

以下の複素数でドモアブルの定理を使いz^413を計算しなさい。
(1)z=(-√3/2)-(1/2)i
(2)z=(√2/2)-(√2/2)i

(1)は(√3/2)-(1/2)i
(2)は(-1/√2)+(1/√2)i

で合ってますでしょうか？
間違っている場合は計算式を教えて頂けるとありがたいです。よろしくお願いします。

回答No.2

jcpmutura · ベストアンサー

合ってます(#1回答の方は間違いです)

(1)
z = (-√3/2) - (1/2)i
= -((√3/2) + (1/2)i)
= -(cos(π/6) + isin(π/6))
より,
z^413 = -(cos(413π/6) + isin(413π/6))
= -(cos(5π/6) + isin(5π/6))
= -(-cos(π/6) + isin(π/6))
= cos(π/6) - isin(π/6)
= (√3/2)-(1/2)i

(2)
z = (√2/2)-(√2/2)i
= cos(π/4) - isin(π/4)
より、
z^413 = cos(413π/4) - isin(413π/4)
= cos(103π+π/4) - isin(103π+π/4)
= cos(5π/4) - isin(5π/4)
= -cos(π/4) + isin(π/4)
= (-1/√2) + (1/√2)i

回答No.1

asuncion

(1)
z = (-√3/2) - (1/2)i
= -((√3/2) + (1/2)i)
= -(cos(π/6) + isin(π/6))
より、
z^413 = -(cos(413π/6) + isin(413π/6))
= -(cos(5π/6) + isin(5π/6))
= -√3/2 - (1/2)i

(2)
z = cos(π/4) - isin(π/4)
より、
z^413 = cos(413π/4) - isin(413π/4)
= cos(π/4) - isin(π/4)
= 1/√2 - (1/√2)i