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How can it be proved that:



Thanks in advance! :-)

Note by Aniruddha Bhattacharjee
8 months ago

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\(i= cos(\frac{\pi}{2}) + isin(\frac{\pi}{2})\implies i=e^{i\frac{\pi}{2}}\implies i^i = (e^{i\frac{\pi}{2}})^{i} = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}}\)

Taking logarithm on both sides we get,

\(ln(i^i) = -\frac{\pi}{2}\)

So finally,

\(\color{red}{-2ln(i^i)=\pi}\) Aditya Sharma · 8 months ago

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@Aditya Sharma Ok.....that was pretty easy....foolish me! Thanks a lot! :-) Aniruddha Bhattacharjee · 8 months ago

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