Let z satisfy $$|z| = 1$$ and $$z = 1 - \bar{z}$$.
Statement 1 : z is a real number.
Statement 2 : Principal argument of z is $$\frac{\pi}{3}$$.

Let $$z = a + bi$$ with $$a, b \in \mathbb{R}$$. The condition $$z = 1 - \bar{z}$$ gives $$a + bi = 1 - (a - bi) = (1-a) + bi$$. Comparing real parts: $$a = 1 - a$$, so $$a = \frac{1}{2}$$.

The condition $$|z| = 1$$ gives $$a^{2} + b^{2} = 1$$, hence $$\frac{1}{4} + b^{2} = 1$$, which yields $$b^{2} = \frac{3}{4}$$, so $$b = \pm\frac{\sqrt{3}}{2}$$. The two solutions are $$z = \frac{1}{2} + \frac{\sqrt{3}}{2}\,i$$ and $$z = \frac{1}{2} - \frac{\sqrt{3}}{2}\,i$$.

Statement 1 claims $$z$$ is a real number. Since both solutions have a nonzero imaginary part ($$b = \pm\frac{\sqrt{3}}{2}$$), Statement 1 is false.

Statement 2 claims the principal argument of $$z$$ is $$\frac{\pi}{3}$$. For $$z = \frac{1}{2} + \frac{\sqrt{3}}{2}\,i$$, the argument is $$\arctan\!\left(\frac{\sqrt{3}/2}{1/2}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}$$. So there exists a valid $$z$$ for which the principal argument is indeed $$\frac{\pi}{3}$$, making Statement 2 true.

Therefore Statement 1 is false and Statement 2 is true.