Let z in mathbb{C} and i=sqrt{-1}. If a, b, c | vedclass

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(A) Given that $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$.$z = \frac{b+ic}{1+a}$.Then $iz = \frac{-c+ib}{1+a}$.Using the property $\frac{1+w}{1-w}$ for $w=iz$:

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$\frac{a+ib}{1+c}$

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(A) Given that $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$.$z = \frac{b+ic}{1+a}$.Then $iz = \frac{-c+ib}{1+a}$.Using the property $\frac{1+w}{1-w}$ for $w=iz$:$\frac{1+iz}{1-iz} = \frac{1 + \frac{-c+ib}{1+a}}{1 - \frac{-c+ib}{1+a}} = \frac{1+a-c+ib}{1+a+c-ib}$.Multiply numerator and denominator by the conjugate of the denominator $(1+a+c)+ib$:$= \frac{((1+a)-c+ib)((1+a)+c+ib)}{(1+a+c)^2+b^2} = \frac{((1+a)+ib)^2 - c^2}{(1+a)^2 + 2c(1+a) + c^2 + b^2}$.Since $a^2+b^2+c^2=1$,we have $b^2+c^2 = 1-a^2$.Denominator $= (1+a)^2 + 2c(1+a) + (1-a^2) = 1+2a+a^2 + 2c(1+a) + 1-a^2 = 2+2a+2c(1+a) = 2(1+a)(1+c)$.Numerator $= (1+a)^2 + 2ib(1+a) - b^2 - c^2 = (1+a)^2 + 2ib(1+a) - (1-a^2) = 1+2a+a^2 + 2ib(1+a) - 1 + a^2 = 2a^2+2a+2ib(1+a) = 2a(a+1) + 2ib(1+a) = 2(a+1)(a+ib)$.Thus,$\frac{1+iz}{1-iz} = \frac{2(1+a)(a+ib)}{2(1+a)(1+c)} = \frac{a+ib}{1+c}$.

Let $z \in \mathbb{C}$ and $i=\sqrt{-1}$. If $a, b, c \in (0,1)$ are such that $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$,then $\frac{1+iz}{1-iz}=$

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