i to the power of i  

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Let's derive ii using very simple algebra. We know that e = -1. We also know that i2 = -1. Combining these facts, we get that e = i2. Bringing both sides to the ith power, we get ei*iπ = i2i ⇒ e = i2i. We can then take the sqrt of both sides, getting that ii = √e = e-π/2. My challenge to you is to derive this without using the identity e. Good Luck!

That one guy that's always on

 
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I've been thinking about this for a day or so, and can't think of a way of doing it without using Euler's identity like you said. You can do it via De Moivre's theorem, but that still requires e = -1. 

I would be interested to see if anyone else comes up with something. 

PlanckTime - Amin

Just as  a suggestion, you can use <sup></sup> in the source code to add exponents. For example, e<sup>i&pi;</sup> returns e.

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Thanks Lucas that's awesome!!

PlanckTime - Amin

 
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