Question Number 68964 by ahmadshah last updated on 17/Sep/19
Commented by mind is power last updated on 17/Sep/19
$${m}+{in}={e}^{{ia}} +{e}^{{ib}} \Rightarrow{m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{2}{imn}={e}^{{i}\mathrm{2}{a}} +{e}^{\mathrm{2}{ib}} +\mathrm{2}{e}^{{i}\left({a}+{b}\right)} \\ $$$$\Rightarrow{Im}\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{2}{imn}\right)={Im}\left({e}^{\mathrm{2}{ia}} +{e}^{\mathrm{2}{ib}} +\mathrm{2}{e}^{{i}\left({a}+{b}\right)} \right) \\ $$$$\Rightarrow\mathrm{2}{mn}={sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{2}{b}\right)+\mathrm{2}{sin}\left({a}+{b}\right) \\ $$$$\Rightarrow\mathrm{2}{mn}=\mathrm{2}{cos}\left({a}−{b}\right){sin}\left({a}+{b}\right)+\mathrm{2}{sin}\left({a}+{b}\right) \\ $$$$\Rightarrow{mn}={sin}\left({a}+{b}\right)\left({cos}\left({a}−{b}\right)+\mathrm{1}\right) \\ $$$$\mid{m}+{in}\mid^{\mathrm{2}} =\mid{e}^{{ia}} +{e}^{{ib}} \mid^{\mathrm{2}} \Rightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\left({cos}\left({a}\right)+{cos}\left({b}\right)\right)^{\mathrm{2}} +\left({sin}\left({a}\right)+{sin}\left({b}\right)\right)^{\mathrm{2}} \\ $$$$\Rightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}{cos}\left({a}\right){cos}\left({b}\right)+\mathrm{2}{sin}\left({a}\right){sin}\left({b}\right)\Rightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+{cos}\left({a}−{b}\right)\right) \\ $$$$\Rightarrow\mathrm{1}+{cos}\left({a}−{b}\right)=\frac{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{mn}={sin}\left({a}+{b}\right)\left(\frac{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow{son}\left({a}+{b}\right)=\frac{\mathrm{2}{mn}}{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$