If z(1 + a) = b + ic and a^2 + b^2 + c^2 = 1, | vedclass

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(A) Given $z = \frac{b + ic}{1 + a}$.Substituting $z$ into the expression $\frac{1 + iz}{1 - iz}$:$\frac{1 + i(\frac{b + ic}{1 + a})}{1 - i(\frac{b +

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

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(A) Given $z = \frac{b + ic}{1 + a}$.Substituting $z$ into the expression $\frac{1 + iz}{1 - iz}$:$\frac{1 + i(\frac{b + ic}{1 + a})}{1 - i(\frac{b + ic}{1 + a})} = \frac{1 + a + ib - c}{1 + a - ib + c} = \frac{(1 + a - c) + ib}{(1 + a + c) - ib}$.Multiplying numerator and denominator by the conjugate of the denominator,$(1 + a + c) + ib$:$= \frac{((1 + a - c) + ib)((1 + a + c) + ib)}{(1 + a + c)^2 + b^2}$.Expanding the numerator:$= \frac{(1 + a)^2 - c^2 + ib(1 + a - c) + ib(1 + a + c) - b^2}{(1 + a + c)^2 + b^2} = \frac{1 + 2a + a^2 - c^2 - b^2 + 2ib(1 + a)}{(1 + a + c)^2 + b^2}$.Since $a^2 + b^2 + c^2 = 1$,we have $a^2 - b^2 - c^2 = 2a^2 - 1$.Numerator $= 1 + 2a + 2a^2 - 1 + 2ib(1 + a) = 2a(1 + a) + 2ib(1 + a) = 2(1 + a)(a + ib)$.Denominator $= (1 + a)^2 + c^2 + 2c(1 + a) + b^2 = (1 + a)^2 + (b^2 + c^2) + 2c(1 + a) = (1 + a)^2 + (1 - a^2) + 2c(1 + a) = (1 + a)^2 + (1 - a)(1 + a) + 2c(1 + a) = (1 + a)(1 + a + 1 - a + 2c) = (1 + a)(2 + 2c) = 2(1 + a)(1 + c)$.Thus,the expression simplifies to $\frac{2(1 + a)(a + ib)}{2(1 + a)(1 + c)} = \frac{a + ib}{1 + c}$.

If $z(1 + a) = b + ic$ and $a^2 + b^2 + c^2 = 1$,then $\frac{1 + iz}{1 - iz} = $

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