Complex Numbers And Quadratic Equations question 435
Question: If $ z(1+a)=b+ic $ and $ a^{2}+b^{2}+c^{2}=1 $ , then $ \frac{1+iz}{1-iz}= $
Options:
A) $ \frac{a+ib}{1+c} $
B) $ \frac{b-ic}{1+a} $
C) $ \frac{a+ic}{1+b} $
D) None of these
Show Answer
Answer:
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} $ .
BETA
