Question Number 110367 by mathdave last updated on 28/Aug/20
Answered by Dwaipayan Shikari last updated on 28/Aug/20
$$\left.\mathrm{1}−\left({log}\left(−\mathrm{1}\right)\right)\right)^{\mathrm{2}} −\mathrm{4}{ilog}\left({i}\right) \\ $$$$\mathrm{1}−\left(\pi{i}\right)^{\mathrm{2}} −\mathrm{4}{i}\left(\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by Aziztisffola last updated on 28/Aug/20
$$\mathrm{1}−\left(\mathrm{2ln}\left(\mathrm{i}\right)\right)^{\mathrm{2}} −\mathrm{4iln}\left(\mathrm{i}\right)=\mathrm{1}−\mathrm{4ln}^{\mathrm{2}} \left(\mathrm{i}\right)−\mathrm{4iln}\left(\mathrm{i}\right) \\ $$$$\:\mathrm{1}−\mathrm{4}\frac{\pi^{\mathrm{2}} \mathrm{i}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }−\mathrm{4i}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}=\mathrm{1}+\pi^{\mathrm{2}} +\mathrm{2}\pi=\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\ $$