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Exercise 1.3 · Q92

Q.Find real values of θ\theta for which 4+3isin⁡θ1−2isin⁡θ\dfrac{4+3i\sin\theta}{1-2i\sin\theta} is purely real.

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Let s=sin⁡θs=\sin\theta. Multiply numerator and denominator of 4+3is1−2is\dfrac{4+3is}{1-2is} by the conjugate 1+2is1+2is: numerator becomes (4+3is)(1+2is)=4+8is+3is+6i2s2=4−6s2+11is(4+3is)(1+2is)=4+8is+3is+6i^2s^2=4-6s^2+11is; denominator becomes (1−2is)(1+2is)=1+4s2(1-2is)(1+2is)=1+4s^2 (real). For the whole fraction to be purely real, the imaginary part of the numerator must vanish (the denominator is alrea …

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