Sigh, complex logarithms??

Yes that is another valid proof but it would still be polite to recognise the valid proof that I had already given you.

How about this :/

\(let\\ \begin{align} x&=ln\; i\\ e^x&=i\\now\\ i&=cos\frac{\pi}{2}+isin\frac{\pi}{2}\\\therefore\\ i&=e^{(\pi/2)i}\\\therefore \\ e^x&=e^{(\pi/2)i}\\\therefore\\ x&=\frac{\pi }{2}\;i \\\therefore\\ ln\;i&=\frac{\sqrt{-\pi^2}}{2} \end{align} \)

I just came up with a proof. Apologies to everyone.

\(\ln(\sqrt{-1})=\dfrac{\ln(-1)}{2}\\ e^{i\pi}+1=0\\ e^{i\pi}=-1\\ i\pi = \ln(-1)\\ \therefore \ln(\sqrt{-1})=\dfrac{\ln(-1)}{2}=\dfrac{i\pi}{2}\)