簡約 (1+i)e^{it}

解

簡約

(1 + i) e^{i t}

(cos (t) - sin (t)) + i (cos (t) + sin (t))

解答ステップ

(1 + i) e^{i t}

e^{i t} = cos (t) + i sin (t)

= (1 + i) (cos (t) + i sin (t))

複素数の算術規則を適用する:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

a = 1, b = 1, c = cos (t), d = sin (t)

= (1 \cdot cos (t) - 1 \cdot sin (t)) + (1 \cdot sin (t) + 1 \cdot cos (t)) i

簡素化

1 \cdot cos (t) - 1 \cdot sin (t) : cos (t) - sin (t)

簡素化

1 \cdot sin (t) + 1 \cdot cos (t) : sin (t) + cos (t)

= (cos (t) - sin (t)) + (sin (t) + cos (t)) i

括弧を削除する:

(a) = a

= cos (t) - sin (t) + (sin (t) + cos (t)) i

標準的な複素数形式で

cos (t) - sin (t) + (sin (t) + cos (t)) i

を書き換える:

(cos (t) - sin (t)) + (sin (t) + cos (t)) i

= (cos (t) - sin (t)) + (sin (t) + cos (t)) i