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How can it be proved that:

-2*ln(i^i)=pi

?

Thanks in advance! :-)

Note by Aniruddha Bhattacharjee 11 months, 3 weeks 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 Narayan Sharma · 11 months, 3 weeks ago

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@Aditya Narayan Sharma – Ok.....that was pretty easy....foolish me! Thanks a lot! :-) – Aniruddha Bhattacharjee · 11 months, 3 weeks ago

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## Comments

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

Taking logarithm on both sides we get,

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

So finally,

\(\color{red}{-2ln(i^i)=\pi}\) – Aditya Narayan Sharma · 11 months, 3 weeks ago

Log in to reply

– Aniruddha Bhattacharjee · 11 months, 3 weeks ago

Ok.....that was pretty easy....foolish me! Thanks a lot! :-)Log in to reply