簡約 (e^{i(k+1)θ}-1)/(e^{iθ)-1}

解

簡約

\frac{e ^{i (k + 1) θ} - 1}{e ^{i θ} - 1}

\frac{1 - cos ( θ ) - cos ( θ + k θ ) + cos ( θ + k θ ) cos ( θ ) + sin ( θ + k θ ) sin ( θ )}{( - 1 + cos ( θ ) ) ^{2} + sin ^{2} ( θ )} + i \frac{sin ( θ ) - sin ( θ + k θ ) + sin ( θ + k θ ) cos ( θ ) - cos ( θ + k θ ) sin ( θ )}{( - 1 + cos ( θ ) ) ^{2} + sin ^{2} ( θ )}

解答ステップ

\frac{e ^{i (k + 1) θ} - 1}{e ^{i θ} - 1}

e^{i (k + 1) θ} - 1 = cos (k θ + θ) + i sin (k θ + θ) - 1

= \frac{cos ( k θ + θ ) + i sin ( k θ + θ ) - 1}{e ^{i θ} - 1}

e^{i θ} - 1 = cos (θ) + i sin (θ) - 1

= \frac{cos ( k θ + θ ) + i sin ( k θ + θ ) - 1}{cos ( θ ) + i sin ( θ ) - 1}

有理化する

\frac{cos ( k θ + θ ) + i sin ( k θ + θ ) - 1}{cos ( θ ) + i sin ( θ ) - 1} : \frac{sin ( k θ + θ ) sin ( θ ) + ( cos ( θ ) - 1 ) ( cos ( k θ + θ ) - 1 ) + i ( sin ( k θ + θ ) ( cos ( θ ) - 1 ) - sin ( θ ) ( cos ( k θ + θ ) - 1 ) )}{sin ^{2} ( θ ) + ( cos ( θ ) - 1 ) ^{2}}

= \frac{( cos ( k θ + θ ) - 1 ) ( cos ( θ ) - 1 ) + sin ( k θ + θ ) sin ( θ ) + i ( sin ( k θ + θ ) ( cos ( θ ) - 1 ) - sin ( θ ) ( cos ( k θ + θ ) - 1 ) )}{( cos ( θ ) - 1 ) ^{2} + sin ^{2} ( θ )}

標準的な複素数形式で

\frac{( cos ( k θ + θ ) - 1 ) ( cos ( θ ) - 1 ) + sin ( k θ + θ ) sin ( θ ) + i ( - sin ( θ ) ( cos ( k θ + θ ) - 1 ) + sin ( k θ + θ ) ( cos ( θ ) - 1 ) )}{( cos ( θ ) - 1 ) ^{2} + sin ^{2} ( θ )}

を書き換える:

\frac{cos ( k θ + θ ) cos ( θ ) - cos ( k θ + θ ) - cos ( θ ) + 1 + sin ( k θ + θ ) sin ( θ )}{sin ^{2} ( θ ) + ( - 1 + cos ( θ ) ) ^{2}} + \frac{- cos ( k θ + θ ) sin ( θ ) + sin ( θ ) + sin ( k θ + θ ) cos ( θ ) - sin ( k θ + θ )}{sin ^{2} ( θ ) + ( - 1 + cos ( θ ) ) ^{2}} i

= \frac{cos ( k θ + θ ) cos ( θ ) - cos ( k θ + θ ) - cos ( θ ) + 1 + sin ( k θ + θ ) sin ( θ )}{sin ^{2} ( θ ) + ( - 1 + cos ( θ ) ) ^{2}} + \frac{- cos ( k θ + θ ) sin ( θ ) + sin ( θ ) + sin ( k θ + θ ) cos ( θ ) - sin ( k θ + θ )}{sin ^{2} ( θ ) + ( - 1 + cos ( θ ) ) ^{2}} i