Complex Numbers And Quadratic Equations question 640
Question: What is $ \frac{{{(1+i)}^{4n+5}}}{{{(1-i)}^{4n+3}}} $ equal to, where n is a natural number and $ i=\sqrt{-1} $ ?
Options:
A) 2
B) $ 2i $
C) $ -2 $
D) i
Show Answer
Answer:
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}}}={{( \frac{1+i}{1-i} )}^{4n+3}}.{{(1+i)}^{2}} $ $ ={{[ \frac{(1+i)(1+i)}{(1-i)(1+i)} ]}^{4n+3}}.(1+i^{2}+2i) $ $ ={{[ \frac{1+i^{2}+2i}{1+1} ]}^{4n+3}}.2i={{(i)}^{4n+3}}.2i=2{{(i)}^{4n+4}} $ $ =2.({i^{4(n+1)}})=2 $
BETA
