Question Number 38957 by Rio Mike last updated on 01/Jul/18
$${solve}\:{for}\:{x}\: \\ $$$${e}^{\mathrm{2}{x}} \:+\:\mathrm{2}{e}^{{x}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 02/Jul/18
$${if}\:\:{x}\:{from}\:{R}\:\:\left({e}\right)\:\Leftrightarrow\:\left({e}^{{x}} \:+\mathrm{1}\right)^{\mathrm{2}} \:=\mathrm{0}\:\Leftrightarrow\:{e}^{{x}} \:+\mathrm{1}\:=\mathrm{0}\: \\ $$$$\:\Leftrightarrow\:{e}^{{x}} =−\mathrm{1}?\Rightarrow{no}?{solution}. \\ $$$${ig}\:{x}\:{from}\:{C}\:\:\left({e}\right)\:\Leftrightarrow\:{e}^{{x}} \:+\mathrm{1}\:=\mathrm{0}\:\Leftrightarrow\:{e}^{{x}} \:=−\mathrm{1}\:\Leftrightarrow \\ $$$${e}^{{x}} \:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\:\Leftrightarrow\:{x}_{{k}} \:={i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:{k}\:\in{Z} \\ $$$${the}\:{complex}\:{x}_{{k}} \:{are}\:{solutions}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
$$\left({e}^{{x}} +\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:{when}\:\:{e}^{{x}\:} =−\mathrm{1} \\ $$$${e}^{{x}} ={e}^{{i}\Pi} \\ $$$${x}={i}\Pi \\ $$$$\left\{{e}^{{i}\Pi} ={cos}\Pi+{isin}\Pi={cos}\Pi+{isin}\Pi=−\mathrm{1}\right\} \\ $$$$ \\ $$
Commented by MJS last updated on 01/Jul/18
$$\mathrm{true}\:\mathrm{but}\:\mathrm{not}\:\mathrm{complete} \\ $$$${x}=\left(\mathrm{2}{z}+\mathrm{1}\right)\pi\mathrm{i};\:{z}\in\mathbb{Z} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jul/18
$${yes}\:{sir}\:{i}\:{put}\:{the}\:{prinicipal}\:{value}\:{only} \\ $$$$ \\ $$