Explanation
A complex number is said to be purely imaginary if $$z+\overline { z } =0$$
If $$z=\dfrac { 3+2isinθ }{ 1−2isinθ } $$
then $$\overline { z } =\overline { \dfrac { 3+2isinθ }{ 1−2isinθ } } =\dfrac { 3-2isinθ }{ 1+2isinθ } $$
$$z+\overline { z } =\dfrac { 3+2isinθ }{ 1−2isinθ } +\dfrac { 3-2isinθ }{ 1+2isinθ } $$
So,$$\dfrac { 3+2isinθ }{ 1−2isinθ } +\dfrac { 3−2isinθ }{ 1+2isinθ } =0$$
$$\dfrac { (3+2isinθ)(1+2isinθ)+(3−2isinθ)(1−2isinθ) }{ (1−2isinθ)(1+2isinθ) } =0$$
$$3+6isinθ+2isinθ−4{ sin }^{ 2 }θ+3−6isinθ−2isinθ−4{ sin }^{ 2 }θ=0$$
$$6−{ 8sin }^{ 2 }θ=0$$
$${ sin }^{ 2 }θ=\dfrac { 3 }{ 4 } $$
$$sinθ=\dfrac { \sqrt { 3 } }{ 2 } =sin\dfrac { \pi }{ 3 } $$
$$θ=nπ+{ (−1) }^{ n }\left( \dfrac { \pi }{ 3 } \right) $$
$$sinθ=\dfrac { \sqrt { 3 } }{ 2 } =sin−\dfrac { \pi }{ 3 } $$
$$θ=nπ+{ (−1) }^{ n }\left( \dfrac { -\pi }{ 3 } \right) =nπ+{ (−1) }^{ n+1 }\left( \dfrac { \pi }{ 3 } \right) $$
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