Complex Numbers And Quadratic Equations question 457
Question: If $ z(1+a)=b+ic $ and $ a^{2}+b^{2}+c^{2}=1 $ ,then $ [(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:
[a] $ \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{2a+2a^{2}+2ib+2iab}{2+2ac+2(a+c)}(\therefore a^{2}+b^{2}+c^{2}=1) $ $ =\frac{a+a^{2}+ib+iab}{1+ac+(a+c)} $ $ =\frac{a(a+1)+ib(a+1)}{(a+1)(c+1)}=\frac{a+ib}{c+1} $
BETA
