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 :
[a]| \[\frac{1+iz}{1-iz}=\frac{1+i(b+ic)/(1+a)}{1-i(b+ic)/(1+a)}\] |
| \[=\frac{1+a-c+ib}{1+a+c-ib}\] |
| \[=\frac{(1+a-c+ib)(1+a+c+ib)}{{{(1+a+c)}^{2}}+{{b}^{2}}}\] |
| \[=\frac{1+2a+{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+2ib+2iab}{1+{{a}^{2}}+{{c}^{2}}+{{b}^{2}}+2ac+2(a+c)}\] |
| \[=\frac{2a+2{{a}^{2}}+2ib+2iab}{2+2ac+2(a+c)}(\therefore {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1)\] |
| \[=\frac{a+{{a}^{2}}+ib+iab}{1+ac+(a+c)}\] |
| \[=\frac{a(a+1)+ib(a+1)}{(a+1)(c+1)}=\frac{a+ib}{c+1}\] |
You need to login to perform this action.
You will be redirected in
3 sec
