# Question-131325

Question Number 131325 by mohammad17 last updated on 03/Feb/21
Commented by mohammad17 last updated on 03/Feb/21
$${find}\:{n} \\$$
Answered by mr W last updated on 04/Feb/21
$${no}\:{solution}\:{for}\:{n}\in{Z}! \\$$$$\\$$$${for}\:{n}\in{C}: \\$$$$\left(\mathrm{1}+{i}\right)^{{n}} =−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{i}\right) \\$$$$\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{8}} \\$$$$\left(\sqrt{\mathrm{2}}{e}^{\frac{\pi{i}}{\mathrm{4}}} \right)^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{8}}{e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}} \\$$$$\left({e}^{\mathrm{ln}\:\sqrt{\mathrm{2}}+\frac{\pi{i}}{\mathrm{4}}} \right)^{{n}−\mathrm{1}} ={e}^{−\mathrm{3ln}\:\mathrm{2}+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}} \\$$$$\left({n}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\frac{\pi{i}}{\mathrm{4}}\right)=−\mathrm{3ln}\:\mathrm{2}+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i} \\$$$$\Rightarrow{n}=\mathrm{1}+\frac{\mathrm{4}\left(−\mathrm{3ln}\:\mathrm{2}+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}\right)}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\pi{i}} \\$$$$\Rightarrow{n}=\mathrm{1}+\frac{\mathrm{4}\left(−\mathrm{3ln}\:\mathrm{2}+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}\right)\left(\mathrm{2}\:\mathrm{ln}\:\mathrm{2}−\pi{i}\right)}{\mathrm{4}\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} } \\$$$$\Rightarrow{n}=\frac{\left(\mathrm{8}{k}+\mathrm{5}\right)\pi^{\mathrm{2}} −\mathrm{20}\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{4}{k}+\mathrm{5}\right)\pi{i}}{\mathrm{4}\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} +\pi^{\mathrm{2}} } \\$$
Commented by MJS_new last updated on 04/Feb/21
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\$$$${n}=\frac{\pi^{\mathrm{2}} \left(\mathrm{8}{k}−\mathrm{3}\right)−\mathrm{20ln}^{\mathrm{2}} \:\mathrm{2}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{2}}+\frac{\mathrm{4}\pi\left(\mathrm{4}{k}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}}{\pi^{\mathrm{2}} +\mathrm{4ln}^{\mathrm{2}} \:\mathrm{2}}\mathrm{i} \\$$$$\mathrm{with}\:{k}\in\mathbb{Z} \\$$
Commented by mr W last updated on 04/Feb/21
$${you}\:{are}\:{right}. \\$$$${thanks}! \\$$