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Question

The imaginary part of $fraction numerator e to the power of i theta end exponent over denominator 1 minus k e to the power of i phi end exponent end fraction$ is-

$\frac{\cos θ - kcos (θ - ϕ)}{1 - 2 kcos ϕ + k^{2}}$
$\frac{\cos θ + kcos (θ - ϕ)}{1 - 2 kcos ϕ + k^{2}}$
$\frac{\sin θ - k \sin (ϕ - θ)}{1 - 2 k \cos ϕ + k^{2}}$
$\frac{\sin θ - k \sin (θ - ϕ)}{1 - 2 k \cos ϕ + k^{2}}$

The correct answer is: $fraction numerator sin theta minus k sin left parenthesis theta minus ϕ right parenthesis over denominator 1 minus 2 k cos ϕ plus k squared end fraction$

]
= $fraction numerator e to the power of i theta end exponent minus k e to the power of i left parenthesis theta minus phi right parenthesis end exponent over denominator 1 minus k left parenthesis e to the power of i phi end exponent plus e to the power of negative i phi end exponent right parenthesis plus k to the power of 2 end exponent end fraction$
= $fraction numerator cos invisible function application theta minus k cos invisible function application left parenthesis theta minus phi right parenthesis plus i left square bracket sin invisible function application theta minus k sin invisible function application left parenthesis theta minus phi right parenthesis right square bracket over denominator 1 minus 2 k cos invisible function application phi plus k to the power of 2 end exponent end fraction$
 Imaginary part = $fraction numerator sin invisible function application theta minus k sin invisible function application left parenthesis theta minus phi right parenthesis over denominator 1 minus 2 k cos invisible function application phi plus k to the power of 2 end exponent end fraction$

Related Questions to study

If the lines $x to the power of 2 end exponent plus 2 x y minus 35 y to the power of 2 end exponent minus 4 x plus 44 y minus 12 equals 0$ and $5 x plus lambda y minus 8 equals 0$ are concurrent, then the value of l is

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D: The square of the distance from (0,0) to the point of intersection of $x to the power of 2 end exponent minus y to the power of 2 end exponent minus x minus 3 y minus 2 equals 0$ Arrange A,B,C,D in ascending order

A: The product of the perpendiculars from $open parentheses 2 comma negative 1 close parentheses text to the lines end text 2 x to the power of 2 end exponent plus 6 x y plus y to the power of 2 end exponent equals 0$
B: The product of perpendiculars from (1,1) to the pair of lines $3 x to the power of 2 end exponent plus 10 x y plus 3 y to the power of 2 end exponent minus 28 x minus 28 y plus 49 equals 0$
C: The product of perpendiculars from (0,0) to pair of lines $2 x to the power of 2 end exponent minus 13 x y minus 7 y to the power of 2 end exponent plus x plus 23 y minus 6 equals 0$
D: The square of the distance from (0,0) to the point of intersection of $x to the power of 2 end exponent minus y to the power of 2 end exponent minus x minus 3 y minus 2 equals 0$ Arrange A,B,C,D in ascending order

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$text If end text 3 x to the power of 2 end exponent minus 11 x y plus 10 y to the power of 2 end exponent minus 7 x plus 13 y plus k equals 0$ denotes a pair of straight lines, then the point of intersection of the lines is

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The equation of the pair of lines through the point (a, b) and parallel to the coordinate axes is

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parallel

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parallel

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