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

solve:(((2+i)^(2020) )/((2−i)^(2019) ))

$${solve}:\frac{\left(\mathrm{2}+{i}\right)^{\mathrm{2020}} }{\left(\mathrm{2}−{i}\right)^{\mathrm{2019}} } \\ $$

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

(2+i)=(√5) e^(itan^(−1) ((1/2)))   (2−i)=(√5)e^(−itan^(−1) ((1/2)))   ((((2+i))/((2−i))))^(2020) (2−i)=e^(4040itan^(−1) ((1/2))) (√5)e^(−itan^(−1) ((1/2))) =(√5)e^(4039itan^(−1) ((1/2)))

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

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