There was something interesting that I found out when looking at Euler's Identity. If e^(pi*i)= -1 and i^2 = -1 then e^(pi*i)has to equal i^2 to get the equation seen here.

[tex]e^{\pi \cdot i} = i^{2}[/tex]

My question is can you in this equation solve for i and if you can what would you get? Will i equal to square root of -1 or will you get something different?

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