Proof?

How can it be proved that:

-2*ln(i^i)=pi

?

Note by Aniruddha Bhattacharjee
2 years, 4 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,

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago