solve-2-i-2020-2-i-2019-

Question Number 129613 by mohammad17 last updated on 16/Jan/21

s o l v e : \frac{{(2 + i)}^{2020}}{{(2 - i)}^{2019}}

Answered by Dwaipayan Shikari last updated on 16/Jan/21

(2 + i) = \sqrt{5} e^{{i t a n}^{- 1} (\frac{1}{2})} (2 - i) = \sqrt{5} e^{- {i t a n}^{- 1} (\frac{1}{2})}

{(\frac{(2 + i)}{(2 - i)})}^{2020} (2 - i) = e^{4040 {i t a n}^{- 1} (\frac{1}{2})} \sqrt{5} e^{- {i t a n}^{- 1} (\frac{1}{2})} = \sqrt{5} e^{4039 {i t a n}^{- 1} (\frac{1}{2})}