Search results

I was trying to solve for x. I was trying to solve for i. i should be the square root of -1 however there should be no real answer to the square root of -1 so the square root of -1 is a different type of number which we call i. What I was trying to figure out is if there was way, through Euler's...

I tried taking logs of both sides and using the property rule. I got stuck at pi*i log e = 2log i and from log e^(pi*i)=log i^2.

I'm not sure how to do logs with complex exponents which is why I asked if anyone knew how and what the answer would be.

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. e^{\pi \cdot i} = i^{2} My question is can you in this equation solve for i and if you can what would you get? Will i...