Answer
(a)=$e^{i(θ1+θ2)}$
(b)=$\frac{1}{(cos(θ)+isin(θ))}$=$\frac{1}{e^{iθ}}$
Work Step by Step
(a) $e^{iθ1}$$\times$$e^{iθ2}$=($cosθ1 + isinθ1 $)($cosθ2 + isinθ2 $)
=($cosθ1\times cosθ2-sinθ1\times sinθ2$)+$i$($sinθ1\times cosθ2-sinθ2\times cosθ1$)
=$cos(θ1+θ2)+isin(θ1+θ2)$
=$e^{i(θ1+θ2)}$
(b) $e^{-iθ}$ = $cos(-θ)+isin(-θ)$
=$cos(θ)-isin(θ)$=($cos(θ)-isin(θ)$)=$\frac{(cos(θ)+isin(θ))}{(cos(θ)+isin(θ))}$ x ($cos(θ)-isin(θ)$)
=$\frac{1}{(cos(θ)+isin(θ))}$=$\frac{1}{e^{iθ}}$
