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Question Number 32159 by rahul 19 last updated on 20/Mar/18
Express the following in a+ib form:  (((cos x+isin x)(cos y+isin y))/((cosa+isin a)(cosb+isinb))).
$$\boldsymbol{{E}}{xpress}\:{the}\:{following}\:{in}\:{a}+{ib}\:{form}: \\ $$$$\frac{\left(\mathrm{cos}\:{x}+{i}\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{y}+{i}\mathrm{sin}\:{y}\right)}{\left({cosa}+{i}\mathrm{sin}\:{a}\right)\left({cosb}+{isinb}\right)}. \\ $$
Commented by abdo imad last updated on 20/Mar/18
E =((e^(ix)  e^(iy) )/(e^(ia)  e^(ib) )) =e^(i(x+y))  e^(−i(a+b)) =e^(i(x+y−a−b))   =cos(x+y−a−b) +i sin(x+y −a −b) .
$${E}\:=\frac{{e}^{{ix}} \:{e}^{{iy}} }{{e}^{{ia}} \:{e}^{{ib}} }\:={e}^{{i}\left({x}+{y}\right)} \:{e}^{−{i}\left({a}+{b}\right)} ={e}^{{i}\left({x}+{y}−{a}−{b}\right)} \\ $$$$={cos}\left({x}+{y}−{a}−{b}\right)\:+{i}\:{sin}\left({x}+{y}\:−{a}\:−{b}\right)\:. \\ $$
Commented by rahul 19 last updated on 20/Mar/18
thank you !
$${thank}\:{you}\:! \\ $$
Answered by Tinkutara last updated on 20/Mar/18
=((e^(ix) e^(iy) )/(e^(ia) e^(ib) ))=e^(i(x+y−a−b))   =cos (x+y−a−b)+isin (x+y−a−b)
$$=\frac{{e}^{{ix}} {e}^{{iy}} }{{e}^{{ia}} {e}^{{ib}} }={e}^{{i}\left({x}+{y}−{a}−{b}\right)} \\ $$$$=\mathrm{cos}\:\left({x}+{y}−{a}−{b}\right)+{i}\mathrm{sin}\:\left({x}+{y}−{a}−{b}\right) \\ $$
Commented by rahul 19 last updated on 20/Mar/18
thank you !
$${thank}\:{you}\:! \\ $$

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