e to the pi times i

e^PiSqrt(e^PiSqrt(e^PiSqrt(...(i)))))) = -1

WHY

euler said so

What is the name of this please.

Now you're just fucking with me.

e^(2(pi)i) = 0. big fucking deal.

If e^(2(pi)i) = 0
Then 1/e^(2(pi)i) = 1/0 OH SHI-

>>5
Which is different from the OP's post. Go away.

>>6
We're already dealing with functions that require four dimensions to graph. 1/0 is the least of our problems.

>>5
No it doesn't, shithead.

e^(2(pi)i) = (e^((pi)i))^2 = (-1)^2 = 1
e^(2(pi)i) = cos(2pi) + i*sin(2pi) = 1

e^(pi*sqrt( -1 )) = -1
e^(pi*sqrt( e^(pi*sqrt( -1 )) )) = e^(pi*sqrt( -1 ))
e^(pi*sqrt( e^(pi*sqrt( e^(pi*sqrt( -1 )) )) )) = e^(pi*sqrt( e^(pi*sqrt( -1 )) )) etc.

Just keep using the first line to replace -1 on both sides.  But don't start with e^(pi*sqrt( i )), it won't work.

Raising a real number to the power of an imaginary number doesn't make intuitive sense, but it makes some sort of sense to take the Taylor series for e^x and proclaim that it applies to all complex numbers instead of just real ones. Take a look at the Taylor expansions of sin(x), cos(x) and e^x, you'll see that e^(ix) = cos(x) + i sin(x). From there, you'll see why e^(i pi) is -1.

>>10
Yes, then you can take this and graph it for all x in the complex plane. You'll see you end up with a unit circle where {x = your degrees = theta} and e^(ix) gives you a complex number that defines the vector from the origin to your point on the unit circle.

vector for a point on a circle in cartesian space = <cos(x), sin(x)>
vector for a point on a circle in complex space = <cos(x), (i)sin(x)>

The complex plane is just like the normal plane except instead of X and Y axis' you have real and imaginary axis'.

I can describe this in more detail if you wish, it's actually a pretty simple concept, I think. I'm just here because /b/ is down, so I don't know if I'll be coming back once it does.

>>5
No, at e^((i)(2pi)) you have a vector on the complex plane going from the origin to positive one on the real axis (it is the same as if you did e^((i)(0)) ). Therefore the value it gives out is '1 + 0i' or just '1'.