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

If \[z(1+a)=b+ic\] and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\], then \[\

1.	If \[z(1+a)=b+ic\] and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\], then \[\frac{1+iz}{1-iz}=\]
A.	\[\frac{a+ib}{1+c}\]
B.	\[\frac{b-ic}{1+a}\]
C.	\[\frac{a+ic}{1+b}\]
D.	None of these
Answer» B. \[\frac{b-ic}{1+a}\]

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

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