If z = x + iy and x^2 + y^2 = 1,then frac{1 + | vedclass

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(B) Given $z = x + iy$ and $x^2 + y^2 = 1$. We know that $|z|^2 = x^2 + y^2 = 1$,so $z\bar{z} = 1$,which implies $\bar{z} = \frac{1}{z}$. Consider the

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$z$

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(B) Given $z = x + iy$ and $x^2 + y^2 = 1$. We know that $|z|^2 = x^2 + y^2 = 1$,so $z\bar{z} = 1$,which implies $\bar{z} = \frac{1}{z}$. Consider the expression $E = \frac{1 + x + iy}{1 + x - iy}$. Substitute $x + iy = z$ and $x - iy = \bar{z}$: $E = \frac{1 + z}{1 + \bar{z}}$. Since $\bar{z} = \frac{1}{z}$,we have: $E = \frac{1 + z}{1 + \frac{1}{z}} = \frac{1 + z}{\frac{z + 1}{z}}$. $E = \frac{(1 + z) \cdot z}{z + 1} = z$. Thus,the correct option is $B$.

If $z = x + iy$ and $x^2 + y^2 = 1$,then $\frac{1 + x + iy}{1 + x - iy} = $

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