A) 2
B) \[2i\]
C) \[-2\]
D) i
Correct Answer: A
Solution :
Given \[\frac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}}\] \[=\frac{{{(1+i)}^{4n+3}}.{{(1+i)}^{2}}}{{{(1-i)}^{4n+3}}}={{\left( \frac{1+i}{1-i} \right)}^{4n+3}}.{{(1+i)}^{2}}\] \[={{\left[ \frac{(1+i)(1+i)}{(1-i)(1+i)} \right]}^{4n+3}}.(1+{{i}^{2}}+2i)\] \[={{\left[ \frac{1+{{i}^{2}}+2i}{1+1} \right]}^{4n+3}}.2i={{(i)}^{4n+3}}.2i=2{{(i)}^{4n+4}}\] \[=2.({{i}^{4(n+1)}})=2\]
You need to login to perform this action.
You will be redirected in
3 sec
