24th February

$a+bi= \sqrt{a^2+b^2} e^{i\tan^{-1}\dfrac{b}{a}}$ Similarly $\log(a+bi)= \log(\sqrt{a^2+b^2}) + i\tan^{-1} (\dfrac{b}{a})$ $\log(1+ae^{ix})= \log(\sqrt{(1+a \cos(x))^2+a^2\sin^2(x)}) +i\tan^{-1} (\dfrac{a\sin(x)}{a\cos(x)+1})$ $\log(1+ae^{ix})=( a\cos(x)-\dfrac{a^2}{2} \cos(2x)+\dfrac{a^3}{3} \cos(3x)-\cdots )+i(a\sin(x)-\dfrac{a^2}{2}\sin(2x)+\dfrac{a^3}{3}\sin(3x)-\cdots$ Matching both sides we get $\tan^{-1} (\dfrac{a\sin(x)}{a\cos(x)+1})= a\sin(x)-\dfrac{a^2}{2}\sin(2x)+\dfrac{a^3}{3}\sin(3x)-\cdots$ Take $a=\dfrac{1}{e}$

$\dfrac{\sin(1)}{e} -\dfrac{\sin(2)}{2e^2} +\dfrac{\sin(3)}{3e^3}-\dfrac{\sin(4)}{4e^4}+\cdots= \tan^{-1}(\dfrac{\sin(1)}{\cos(1)+e})$ Answer is $\boxed{a=1}$