If the real part of the complex number $$z=3+2icos$$⁡θ$$1$$−$$3icos$$⁡θ,θ∈$$0$$,π$$2$$ is zero, then the value of $$sin2$$⁡$$3$$θ$$+cos2$$⁡θ is equal to

$$z=3+2icos$$⁡θ$$1$$−$$3icos$$⁡θ$$=3+2icos$$⁡θ$$1+3icos$$⁡θ$$1$$−$$3icos$$⁡θ$$1+3icos$$⁡θ Re⁡$$(z)=3$$−$$6cos2$$⁡θ$$1+9cos2$$⁡θ$$=0$$ ⇒θ$$=$$π$$4$$ Hence, $$sin2$$⁡$$3$$θ$$+cos2$$⁡θ$$=1$$

Questions

$If the real part of the complex number z = \frac{3 + 2 i \cos θ}{1 - 3 i \cos θ}, θ \in (0, \frac{π}{2}) is zero, then the value of \sin^{2} 3 θ + \cos^{2} θ is equal to$

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detailed solution

$z = \frac{3 + 2 i \cos θ}{1 - 3 i \cos θ} = \frac{(3 + 2 i \cos θ) (1 + 3 i \cos θ)}{(1 - 3 i \cos θ) (1 + 3 i \cos θ)} Re (z) = \frac{3 - 6 \cos^{2} θ}{1 + 9 \cos^{2} θ} = 0 \Rightarrow θ = \frac{π}{4} Hence, \sin^{2} 3 θ + \cos^{2} θ = 1$