Question Number 37948 by Fawomath last updated on 19/Jun/18
$${If}\:{x}\:\in\mathbb{R} \\ $$$${show}\:{that}\:\left(\mathrm{2}+{i}\right){e}^{\left(\mathrm{1}+\mathrm{3}{i}\right)} +\left(\mathrm{2}−{i}\right){e}^{\left(\mathrm{1}−\mathrm{3}{i}\right)} \:{is}\:{also}\:{real}. \\ $$
Commented by prof Abdo imad last updated on 19/Jun/18
$${let}\:{prove}\:{first}\:{that}\:{conj}\left(\:{e}^{{a}+{ib}} \right)\:={e}^{{a}−{ib}} \\ $$$${if}\:{a}\:{and}\:{b}\:{reals}\:\:\left({conj}\left({z}\right)=\overset{−} {{z}}\right) \\ $$$${we}\:{have}\:{e}^{{a}\:+{ib}} ={e}^{{a}} \left(\:{cosb}\:\:+{i}\:{sinb}\:\right)\Rightarrow \\ $$$${vonj}\left({e}^{{a}+{ib}} \right)\:={e}^{{a}} \left({cosb}\:−{isinb}\right)={e}^{{a}−{ib}} \:\:{let} \\ $$$${z}\:=\left(\mathrm{2}+{i}\right)\:{e}^{\mathrm{1}+\mathrm{3}{i}} \:\Rightarrow\overset{−} {{z}}=\left(\mathrm{2}−{i}\right){e}^{\mathrm{1}−\mathrm{3}{i}} \:\Rightarrow \\ $$$$\left(\mathrm{2}+{i}\right){e}^{\mathrm{1}+\mathrm{3}{i}} \:+\left(\mathrm{2}−{i}\right){e}^{\mathrm{1}−\mathrm{3}{i}} \:={z}\:+\overset{−} {{z}}=\mathrm{2}{Re}\left({z}\right)\:\in\:{R}. \\ $$