Find real - such that 1 - i cos-1 - 2 i cos- is purely imaginary

The correct option is D n π±π/4 Given, #160; 1+icosθ #160;/ 1 2icosθ #160; =( 1+icosθ #160; #160;) ( 1+2icosθ #160; #160;) #160;/ 1+4co ...

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Standard XII

Mathematics

Domain

Find real θ...

Question

Find real $θ$ such that $\frac{1 + i cos θ}{1 - 2 i cos θ}$ is purely imaginary

n π \pm \frac{π}{2}

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n π \pm \frac{π}{6}

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n π \pm \frac{π}{3}

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n π \pm \frac{π}{4}

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Solution

The correct option is D $n π \pm \frac{π}{4}$
Given, $\frac{1 + i cos θ}{1 - 2 i cos θ} = \frac{(1 + i cos θ) (1 + 2 i cos θ)}{1 + 4 {cos}^{2} θ}$
$= \frac{1 + 2 i cos θ + i cos θ - 2 {cos}^{2} θ}{1 + 4 {cos}^{2} θ}$
$= \frac{1 - 2 {cos}^{2} θ + 3 i cos θ}{1 + 4 {cos}^{2} θ}$
$z$ is purely imaginary when $1 - 2 {cos}^{2} θ = 0$
$\Rightarrow cos θ = \pm \frac{1}{\sqrt{2}}$
$θ = n π \pm \frac{π}{4}$
Ans: D

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Find real θ such that 1+icosθ1−2icosθ is purely imaginary

The correct option is D nπ±π4Given,1+icosθ1−2icosθ=(1+icosθ)(1+2icosθ)1+4cos2θ=1+2icosθ+icosθ−2cos2θ1+4cos2θ=1−2cos2θ+3icosθ1+4cos2θz is purely imaginary when 1−2cos2θ=0⇒cosθ=±1√2θ=nπ±π4Ans: D

Find real $θ$ such that $\frac{1 + i cos θ}{1 - 2 i cos θ}$ is purely imaginary