What is (e^{i(k+1)θ}-1)/(e^{iθ)-1} ?

The solution to (e^{i(k+1)θ}-1)/(e^{iθ)-1} is

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simplify (e^{i(k+1)θ}-1)/(e^{iθ)-1}

simplify

\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 )}

Solution steps

\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}

Rationalize

\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 )}

Rewrite

\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 )}

in standard complex form:

\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

(3 - 3 i)^{7}

(e - f k) (e - e f k + f k)

\frac{248 ( s + 5 )}{s ( s + 4 ) ( s ^{2} + 4 )}

(x^{2} - 2 x + 9)^{7}

(v + v^{2} + 1)^{2}

What is (e^{i(k+1)θ}-1)/(e^{iθ)-1} ?
The solution to (e^{i(k+1)θ}-1)/(e^{iθ)-1} is