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How can it be proved that:

-2*ln(i^i)=pi

?

Thanks in advance! :-)

Note by Aniruddha Bhattacharjee 1 year, 10 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,

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

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Ok.....that was pretty easy....foolish me! Thanks a lot! :-)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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,

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

Log in to reply

Ok.....that was pretty easy....foolish me! Thanks a lot! :-)

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