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Question Number 110367 by mathdave last updated on 28/Aug/20

Answered by Dwaipayan Shikari last updated on 28/Aug/20

1−(log(−1)))^2 −4ilog(i)  1−(πi)^2 −4i(((πi)/2))  (π+1)^2

$$\left.\mathrm{1}−\left({log}\left(−\mathrm{1}\right)\right)\right)^{\mathrm{2}} −\mathrm{4}{ilog}\left({i}\right) \\ $$$$\mathrm{1}−\left(\pi{i}\right)^{\mathrm{2}} −\mathrm{4}{i}\left(\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Answered by Aziztisffola last updated on 28/Aug/20

1−(2ln(i))^2 −4iln(i)=1−4ln^2 (i)−4iln(i)   1−4((π^2 i^2 )/2^2 )−4i^2 (π/2)=1+π^2 +2π=(π+1)^2

$$\mathrm{1}−\left(\mathrm{2ln}\left(\mathrm{i}\right)\right)^{\mathrm{2}} −\mathrm{4iln}\left(\mathrm{i}\right)=\mathrm{1}−\mathrm{4ln}^{\mathrm{2}} \left(\mathrm{i}\right)−\mathrm{4iln}\left(\mathrm{i}\right) \\ $$$$\:\mathrm{1}−\mathrm{4}\frac{\pi^{\mathrm{2}} \mathrm{i}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }−\mathrm{4i}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}=\mathrm{1}+\pi^{\mathrm{2}} +\mathrm{2}\pi=\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\ $$

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