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Find $ such that (3 + 2i sin $)/ (1- 2i sin$) is purely imaginary number ?Answer with solution please .
I am replacing $ with x.For the expression to be purely imaginary,(3 + 2i sinx )/ (1 - 2i sinx) = Bi for some real number B.=> 3 + 2i sin x = 2B sin x + BiComparing the real and imaginary parts of LHS and RHS, we get3 = 2B sin x; ……………………………(i)2sinx = B………………………………..(ii)Using the value of B obtained from (ii) in (i),3 = 4 (sin x) ^2=> (sin x) ^2 = 3/4=> sin x = +sqrt(3)/2 , -sqrt(3)/2=> x = Pi/3 + 2nPi, 2Pi/3 + 2nPi, -Pi/3 + 2nPi, 4Pi/3 + 2nPi, 5Pi/3 + 2nPi(Where n is any integer)If the answers are asked in the range [0,2pi], the solutions are:Pi/3, 2Pi/3, 4Pi/3, 5Pi/3
I am replacing $ with x.
For the expression to be purely imaginary,
(3 + 2i sinx )/ (1 - 2i sinx) = Bi for some real number B.
=> 3 + 2i sin x = 2B sin x + Bi
Comparing the real and imaginary parts of LHS and RHS, we get
3 = 2B sin x; ……………………………(i)
2sinx = B………………………………..(ii)
Using the value of B obtained from (ii) in (i),
3 = 4 (sin x) ^2
=> (sin x) ^2 = 3/4
=> sin x = +sqrt(3)/2 , -sqrt(3)/2
=> x = Pi/3 + 2nPi, 2Pi/3 + 2nPi, -Pi/3 + 2nPi, 4Pi/3 + 2nPi, 5Pi/3 + 2nPi
(Where n is any integer)
If the answers are asked in the range [0,2pi], the solutions are:
Pi/3, 2Pi/3, 4Pi/3, 5Pi/3
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