If 3 + i s i n θ 4 - i c o s θ , θ ∈ 0 ,2 π , is a real number, then an argument of s i n θ + i c o s θ is

z = 3 + i sin θ 4 - i cos θ × 4 + i cos θ 4 + i cos θ Im ( z ) = 3 cos θ + 4 sin θ 16 + cos 2 θ As z is purely real ⇒ 3 cos θ + 4 sin θ = 0 ⇒ tan θ = - 3 4 A r g sin θ + i cos θ = π + tan - 1 cos θ sin θ = π + tan - 1 - 4 3 = π - tan - 1 4 3

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MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Medium2 minPYQ_2020

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If $\frac{3 + i s i n θ}{4 - i c o s θ}, θ \in [0,2 π],$ is a real number, then an argument of $s i n θ + i c o s θ$ is

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Solution:

$z = \frac{(3 + i \sin θ)}{(4 - i \cos θ)} \times \frac{(4 + i \cos θ)}{(4 + i \cos θ)}$

$Im (z) = \frac{3 \cos θ + 4 \sin θ}{16 + \cos^{2} θ}$
As $z$ is purely real $\Rightarrow 3 \cos θ + 4 \sin θ = 0 \Rightarrow \tan θ = - \frac{3}{4}$
$A r g (\sin θ + i \cos θ) = π + \tan^{- 1} (\frac{\cos θ}{\sin θ}) = π + \tan^{- 1} (- \frac{4}{3}) = π - \tan^{- 1} (\frac{4}{3})$

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2020

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