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Question Number 139164 by mathlove last updated on 23/Apr/21

	Answered by qaz last updated on 25/Apr/21
	$ln(1−ae^(ix) ) =ln[1−a(cos x+isin x)] =ln((√((1−acos x)^2 +(asin x)^2 ))e^(itan^(−1) ((−asin x)/(1−acos x))) ) =(1/2)ln(1−2acos x+a^2 )+itan^(−1) ((asin x)/(acos x−1))...................≪1≫ ln(1−ae^(ix) )=−Σ_(n=1) ^∞ (a^n /n)e^(inx) =−Σ_(n=1) ^∞ (a^n /n)(cos nx+isin nx)......≪2≫ ≪1≫&≪2≫ ⇒ln(1−2acos x+a^2 )=−2Σ_(n=1) ^∞ (a^n /n)cos nx −−−−−−−−−−−−−−−−−−−− I=−2Σ_(n=1) ^∞ (a^n /n)∫_0 ^π cos (nx)cos^2 (x/2)dx =−Σ_(n=1) ^∞ (a^n /n)∫_0 ^π cos (nx)(cos x+1)dx =−Σ_(n=1) ^∞ (a^n /n){((sin (nx))/n)∣_0 ^π +(1/2)∫_0 ^π cos [(n+1)x]+cos [(n−1)x]dx} =−Σ_(n=1) ^∞ (a^n /n){sin (nπ)+(1/2)[((sin (n+1)x)/(n+1))+((sin (n−1)x)/(n−1))]_0 ^π } =−Σ_(n=1) ^∞ (a^n /n){sin (nπ)+(1/2)[((sin (n+1)π)/(n+1))+((sin (n−1)π)/(n−1))]} =0$
	$l n (1 - {a e}^{i x})$ $= l n [1 - a (\cos x + i \sin x)]$ $= l n (\sqrt{{(1 - a \cos x)}^{2} + {(a \sin x)}^{2}} e^{i \tan^{- 1} \frac{- a \sin x}{1 - a \cos x}})$ $= \frac{1}{2} l n (1 - 2 a \cos x + a^{2}) + i \tan^{- 1} \frac{a \sin x}{a \cos x - 1} . . . . . . . . . . . . . . . . . . . ≪ 1 ≫$ $l n (1 - {a e}^{i x}) = - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} e^{i n x} = - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} (\cos n x + i \sin n x) . . . . . . ≪ 2 ≫$ $≪ 1 ≫ & ≪ 2 ≫$ $\Rightarrow l n (1 - 2 a \cos x + a^{2}) = - 2 \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} \cos n x$ $- - - - - - - - - - - - - - - - - - - -$ $I = - 2 \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} \int_{0}^{π} \cos (n x) \cos^{2} \frac{x}{2} d x$ $= - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} \int_{0}^{π} \cos (n x) (\cos x + 1) d x$ $= - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} {\frac{\sin (n x)}{n} ∣_{0}^{π} + \frac{1}{2} \int_{0}^{π} \cos [(n + 1) x] + \cos [(n - 1) x] d x}$ $= - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} {\sin (n π) + \frac{1}{2} {[\frac{\sin (n + 1) x}{n + 1} + \frac{\sin (n - 1) x}{n - 1}]}_{0}^{π}}$ $= - \underset{n = 1}{\sum^{\infty}} \frac{a^{n}}{n} {\sin (n π) + \frac{1}{2} [\frac{\sin (n + 1) π}{n + 1} + \frac{\sin (n - 1) π}{n - 1}]}$ $= 0$
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