Having trouble understanding why the dot product of e^(pi*i) and the conjugate of its derivative does not equal 0? In my attempt to understand where Euler's identity ($e^{it}=\cos(t)+i\cdot \sin(t)$) I am using the fact that the derivative of $e^{it}$ is $ie^{it}$ which is perpendicular to $e^{it}$ at all $t$ (which is why it forms a circle). However, I can't seem to verify this using Mathematica. 
I read elsewhere that the dot product of continuous functions is $$\frac{1}{T}\int_T{f(t)\overline{g(t)}dt}.$$ So I tried typing this into Mathematica as:
p[t_]:=Exp[I t]
v[t_]:=D[p[t],t]
(1 / (2 Pi)) Integrate[p[t] Conjugate[v[t]], {t, 0,2Pi}]

Which results in $-i$, not $0$ as I was expecting. However, if I do not take the conjugate of $v(t)$, then it does result in $0$. What am I missing here? 
 A: The inner product/dot product for two complex valued, continuous functions, on the interval $[0,T]$ is given by:
$$
\frac{1}{T}\int_0^Tf(t)\overline{g(t)} \mathrm{d}t
$$
Here, the product in the integral is given by the product of complex numbers. If we write:
$$
f(t) = u(t) + iv(t)
$$
$$
g(t) =  p(t) + iq(t) 
$$
Then, we are going to compute the integral:
$$
\int_0^T (u(t) + iv(t))\overline{(p(t) + iq(t))} dt = \int_0^T(u(t) + i v(t)) (p(t) - iq(t))dt $$
$$
 \int_0^T\underbrace{[u(t) p(t) + v(t)q(t)]}_{\text{Real Part}} + \underbrace{[p(t)v(t) -  u(t)q(t)]}_{\text{Imaginary Part}}i
 dt$$
In particular, let $f(t) = e^{it}$, $g(t) = ie^{it}$. If you calculate the above, you'll see that the integrand doesn't vanish.Thus, these functions are not orthogonal because they use the product of complex numbers in the integral.
However, at every $t$, you get vectors in $\mathbb{R}^2$. These vectors (not functions) have a dot product, given by the standard dot product:
$$
d(t) = f(t) \cdot g(t) = u(t)p(t) + v(t)q(t)
$$
In this case, you'll se the above quantity $d(t)$ vanish for all $t$. However, $d(t)$ isn't the quantity we have in the integral, that is the product of complex numbers. This has a different geometric interpretation: while the dot product vanishes for two orthogonal vectors (this is basically the definition of the dot product and of orthogonal), the product of complex numbers does not! 
A: By this time (based on other comments) you may be cluing into the fact that there are many kinds of vector spaces, not all of which have vectors that can easily be drawn as arrows with a finite number of real coordinates.
There are therefore also many kinds of inner products.
The inner product of continuous functions that you got from
Understanding dot product of continuous functions
is based on the idea that the entire definition of the function over its entire domain is a single vector. You can't describe one of these vectors just by giving an $x$ coordinate and a $y$ coordinate. Indeed any finite number of coordinates is not enough in general.
You appear to be trying to treat a complex number as a vector, which makes some intuitive sense if you think of plotting a complex number on a plane with two coordinates, one for  the real part and one for the imaginary part, writing the number as $x + iy.$
The complex numbers you are interested in happen to be functions of a parameter $t,$
but a single value of $t$ gives you a single vector as a result; you do not have a vector corresponding to the entire definition of the function.
The kind of orthogonality you are going for is exemplified by the two formulas for your two vectors:
$$ e^{it} \qquad \text{and} \qquad ie^{it}. $$
Notice that the only thing different about the second formula is the extra factor of $i.$ Multiplying a complex number by $\pm i$ "rotates" it $90$ degrees (in your visualization of the complex plane); if two numbers are orthogonal (in the sense you are looking for) then the ratio of the two numbers is some real multiple of $i.$
That is, if $w$ and $z$ are complex numbers, orthogonal in the sense you want,
then
$$  \frac wz = ir \quad \text{where $i$ is real.}$$
This definition is a bit awkward (the "where $i$ is real" part), but we can use the fact that $ir + \overline{ir} = 0$;
we can say $w$ and $z$ are orthogonal if
$$ \frac wz + \frac {\overline w}{\overline z} = 0. $$
This definition does not work if $z=0,$ but if we multiply all terms by 
$z\overline z$ then we get the equation
$$ w{\overline z} + {\overline w}z = 0. $$
Note that if $w = a + ib$ and $z = c + id$ then
$$ \frac12 \left(w{\overline z} + {\overline w}z\right) = ac + bd, $$
which is what you might want for an inner product.
