Let z satisfy |z|=1, z=1-bar{z} and operatorn | vedclass

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(C) Given $z = x + iy$. Since $|z| = 1$, we have $x^2 + y^2 = 1$.From $z = 1 - \bar{z}$, we get $x + iy = 1 - (x - iy) = 1 - x + iy$.Comparing real pa

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Statement-$I$ is false, Statement-$II$ is true

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(C) Given $z = x + iy$. Since $|z| = 1$, we have $x^2 + y^2 = 1$.From $z = 1 - \bar{z}$, we get $x + iy = 1 - (x - iy) = 1 - x + iy$.Comparing real parts, $x = 1 - x$, which implies $2x = 1$, so $x = \frac{1}{2}$.Since $x^2 + y^2 = 1$, we have $(\frac{1}{2})^2 + y^2 = 1$, so $y^2 = 1 - \frac{1}{4} = \frac{3}{4}$.Given $\operatorname{Im}(z) > 0$, we have $y = \frac{\sqrt{3}}{2}$.Thus, $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.Since $z$ has an imaginary part, Statement-$I$ is false.The argument of $z$ is $	heta = 	an^{-1}(\frac{\sqrt{3}/2}{1/2}) = 	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Thus, Statement-$II$ is true.

Let $z$ satisfy $|z|=1$, $z=1-\bar{z}$ and $\operatorname{Im}(z) > 0$.
Statement-$I$: $z$ is a real number.
Statement-$II$: Principal argument of $z$ is $\frac{\pi}{3}$.
Then

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Let $z$ satisfy $|z|=1$, $z=1-\bar{z}$ and $\operatorname{Im}(z) > 0$.Statement-$I$: $z$ is a real number.Statement-$II$: Principal argument of $z$ is $\frac{\pi}{3}$.Then