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](Q139275.png)
$${ln}\left(\mathrm{1}−{ae}^{{ix}} \right) \\ $$$$={ln}\left[\mathrm{1}−{a}\left(\mathrm{cos}\:{x}+{i}\mathrm{sin}\:{x}\right)\right] \\ $$$$={ln}\left(\sqrt{\left(\mathrm{1}−{a}\mathrm{cos}\:{x}\right)^{\mathrm{2}} +\left({a}\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{e}^{{i}\mathrm{tan}^{−\mathrm{1}} \frac{−{a}\mathrm{sin}\:{x}}{\mathrm{1}−{a}\mathrm{cos}\:{x}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:{x}+{a}^{\mathrm{2}} \right)+{i}\mathrm{tan}^{−\mathrm{1}} \frac{{a}\mathrm{sin}\:{x}}{{a}\mathrm{cos}\:{x}−\mathrm{1}}...................\ll\mathrm{1}\gg \\ $$$${ln}\left(\mathrm{1}−{ae}^{{ix}} \right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}{e}^{{inx}} =−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\left(\mathrm{cos}\:{nx}+{i}\mathrm{sin}\:{nx}\right)......\ll\mathrm{2}\gg \\ $$$$\ll\mathrm{1}\gg\&\ll\mathrm{2}\gg \\ $$$$\Rightarrow{ln}\left(\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:{x}+{a}^{\mathrm{2}} \right)=−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\mathrm{cos}\:{nx} \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$${I}=−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\:\left({nx}\right)\mathrm{cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\:\left({nx}\right)\left(\mathrm{cos}\:{x}+\mathrm{1}\right){dx} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\left\{\frac{\mathrm{sin}\:\left({nx}\right)}{{n}}\mid_{\mathrm{0}} ^{\pi} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}\:\left[\left({n}+\mathrm{1}\right){x}\right]+\mathrm{cos}\:\left[\left({n}−\mathrm{1}\right){x}\right]{dx}\right\} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\left\{\mathrm{sin}\:\left({n}\pi\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{sin}\:\left({n}+\mathrm{1}\right){x}}{{n}+\mathrm{1}}+\frac{\mathrm{sin}\:\left({n}−\mathrm{1}\right){x}}{{n}−\mathrm{1}}\right]_{\mathrm{0}} ^{\pi} \right\} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}^{{n}} }{{n}}\left\{\mathrm{sin}\:\left({n}\pi\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{sin}\:\left({n}+\mathrm{1}\right)\pi}{{n}+\mathrm{1}}+\frac{\mathrm{sin}\:\left({n}−\mathrm{1}\right)\pi}{{n}−\mathrm{1}}\right]\right\} \\ $$$$=\mathrm{0} \\ $$