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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

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

Answered by Aziztisffola last updated on 28/Aug/20

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

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