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

$$solve:\frac{{(2+i)}^{2020}}{{(2-i)}^{2019}}$$

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

$$(2+i)=\sqrt{5}{e}^{{itan}^{-1}\left(\frac{1}{2}\right)}(2-i)=\sqrt{5}{e}^{-{itan}^{-1}\left(\frac{1}{2}\right)}$$$${\left(\frac{(2+i)}{(2-i)}\right)}^{2020}(2-i)=e^{4040{itan}^{-1}\left(\frac{1}{2}\right)}\sqrt{5}{e}^{-{itan}^{-1}\left(\frac{1}{2}\right)}=\sqrt{5}{e}^{4039{itan}^{-1}\left(\frac{1}{2}\right)}$$

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