I’ve come across many questions from students regarding complex numbers, and here is one thats quite fun to deal with!

$\sqrt{-1}^{\sqrt{-1}}$

Now this is really interesting, but can be solved with a bit of manipulation. We all know this is akin to $i^i$

We know that $\sqrt{-1}=e^{i\frac{\pi}{2}}$

$\sqrt{-1}^{\sqrt{-1}}$

$= e^{i\frac{\pi}{2}\sqrt{-1}}$

$= e^{i\frac{\pi}{2}i}$

$= e^{i^2\frac{\pi}{2}}$

$= e^{-\frac{\pi}{2}}$

Now our end result is a real number!

Like mentioned with regards to Euler’s Identity, it is really very amazing how complex numbers actually work!

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