if z=(((3+i sin θ)/(4−i cos θ)))is purely real and (Π/2)<θ<Π then find arg(sin θ +i cos θ)?


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Question Number 145763 by gsk2684 last updated on 07/Jul/21

$i f z = (\frac{3 + i \sin θ}{4 - i \cos θ}) i s p u r e l y r e a l a n d$ $\frac{Π}{2} < θ < Π t h e n f i n d a r g (\sin θ + i \cos θ) ?$
	Answered by mathmax by abdo last updated on 08/Jul/21

	$z real \Rightarrow z = \bar{z} \Rightarrow \frac{3 + isin θ}{4 - icos θ} = \frac{3 - isin θ}{4 + icos θ} \Rightarrow$ $(3 + isin θ) (4 + icos θ) = (3 - isin θ) (4 - icos θ) \Rightarrow$ $12 + 3 icos θ + 4 isin θ - \sin θ \cos θ = 12 - 3 icos θ - 4 isin θ - \sin θ \cos θ \Rightarrow$ $6 icos θ + 8 isin θ = 0 \Rightarrow 2 i (3 \cos θ + 4 \sin θ) = 0 \Rightarrow 3 \cos θ + 4 \sin θ = 0 \Rightarrow$ $5 (\frac{3}{5} \cos θ + \frac{4}{5} \sin θ) = 0 \Rightarrow 5 \cos (θ - α) = 0 (e) with \cos α = \frac{3}{5} and \sin α = \frac{4}{5}$ $\Rightarrow \tan α = \frac{4}{3} \Rightarrow α = \arctan (\frac{4}{3})$ $(e) \Rightarrow θ - α = \frac{π}{2} + k π (k \in Z)$ $\sin θ + icos θ = \sin (α + \frac{π}{2} + k π) + icos (α + \frac{π}{2} + k π)$ $= \cos (α + k π) - isin (α + k π)$ $= {(- 1)}^{k} \cos α - i {(- 1)}^{k} \sin α$ $= {(- 1)}^{k} e^{- i α} = {(- 1)}^{k} e^{- iarctan (\frac{4}{3})}$ $for \frac{π}{2} < θ < π we solve \frac{π}{2} < α + \frac{π}{2} + k π < π . . . . (findk)$
	Commented by gsk2684 last updated on 08/Jul/21

	$t h a n k y o u$
	Commented by mathmax by abdo last updated on 09/Jul/21

	$you are welcome$
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