5+isin⁡θ5−3isin⁡θ  is a real number when

5+isin⁡θ5−3isin⁡θ is a real number⇔ 5+isin⁡θ5−3isin⁡θ=5−isin⁡θ5+3isin⁡θ⇔ (5+isin⁡θ)(5+3isin⁡θ)=(5−3isin⁡θ)(5−isin⁡θ)=25−20isin⁡θ−3sin2⁡θ⇔40isin⁡θ=0⇔sin⁡θ=0This is possible when θ=-π