A) \[\frac{a+ib}{1+c}\]
B) \[\frac{b-ic}{1+a}\]
C) \[\frac{a+ic}{1+b}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{1+iz}{1-iz}=\frac{1+i(b+ic)/(1+a)}{1-i(b+ic)/(1+a)}=\frac{1+a-c+ib}{1+a+c-ib}\] \[=\frac{(1+a-c+ib)(1+a+c+ib)}{{{(1+a+c)}^{2}}+{{b}^{2}}}\] \[=\frac{1+2a+{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+2ib+2iab)}{1+{{a}^{2}}+{{c}^{2}}+{{b}^{2}}+2ac+2(a+c)}\] = \[\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2a+{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+2ib(1+a)}{1+1+2ac+2(a+c)}\] \[=\frac{2a(a+1)+2ib(1+a)}{2(1+a)(1+c)}=\frac{a+ib}{1+c}\].
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