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Find real values of θ for which (4+3isinθ1-2isinθ) is purely real.

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प्रश्न

Find real values of θ for which `((4 + 3"i" sintheta)/(1 - 2"i" sin theta))` is purely real.

योग
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उत्तर

Let z = `(4 + 3"i" sintheta)/(1 - 2"i" sintheta)`

= `(4 + 3"i" sintheta)/(1 - 2"i" sintheta) xx (1 + 2"i" sintheta)/(1 + 2"i" sintheta)`

= `(4 + 8"i" sintheta + 3"i" sintheta + 6"i"^2 sin^2theta)/(1 - 4"i"^2 sin^2theta)`

= `(4 + (11 sintheta)"i" - 6 sin^2theta)/(1 + 4 sin^2theta)`    ...[∵ i2 = – 1]

= `((4 - 6 sin^2theta) + (11 sintheta)"i")/(1 + 4 sin^2theta)`

∴ z = `((4 - 6 sin^2theta)/(1 + 4 sin^2theta)) + ((11 sintheta)/(1 + 4 sin^2theta))"i"`

Since z is purely real, Im(z) = 0

∴ `(11 sintheta)/(1 + 4 sin^2theta)` = 0

∴ sin θ = 0 = sin nπ, where n ∈ Z

∴ θ = nπ, where n ∈ Z.

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अध्याय 1: Complex Numbers - Exercise 1.3 [पृष्ठ १५]

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