Why does $\int_C\frac{dz}{z}=2\pi i$? I was thinking that I don't really understand why
$$\int_{|z|=1}\frac{dz}{z}=2\pi i$$
I do understand how to calculate it and it's importance for residues, Cauchy etc... The question is that, if $f(z)=z^{-1}$, then $f(z)=\overline{z}$ when $|z|=1$, somehow there must be an asymmetry in $f(z)$ along $|z|=1$ in order for the integral to be $2\pi$ units up in the imaginary axis. If $C$ is the unit circle traversed in the natural (anticlockwise) direction at "1 m/s speed"(i.e. $e^{i\theta}$ for $\theta\in[0,2\pi]$) then $f(C)$ is again the unit circle traversed at "1 m/s" speed but in a clockwise manner. Then again, I don't see any asymmetry: you will pass both by $z$ and $-z$ and those should cancel giving $0$ as the value of the integral.
I tried to think of the line integral in terms of area: $ie^{i\theta}$ is the tangent vector to the curve at $z=e^{i\theta}$ and an infinitesimal triangle (or circle sector) should have area $\frac{1}{2}ie^{i\theta}e^{i\theta}$. Couldn't get any further because what meaning does "complex number value area" mean? Is there some interpretation? I am not convinced that the $\gamma'(t)$ factor appearing in the calculations really makes the difference (or how it would).
Just to add to my controversy, the function $f(z)=z^{-2}$ takes the unit circle into the unit circle but "changes speeds" by which I mean that it gives somehow "more weight" to some points (it stretches the upper semicircle into the whole circle)
Maybe the question sounds stupid and I hope it does have a simple explanation.
Thanks! 
 A: Your first argument applies to a related, but different integral, where we really do sum up the values of $f(e^{i\theta})$ over time and find a cancellation between 
$z$ and $-z$:
$$\int_{0}^{2\pi} f(e^{i\theta})\,d\theta = 0.$$
Note, however, that this integral is not of the desired form: the argument $e^{i\theta}$ to the function $f$ is different from the variable we are integrating with $\theta$. The intuition behind this integral is taking an "average" of $\frac{1}z$ over the unit circle, but line integrals are up to something else: they are better thought of as measuring something more like the physical quantity of "how much energy does one gain when pushing an object counterclockwise around the circle against a force acting clockwise?" rather than "at a random angle, what is the expectation of my position?" - and note that this is captured in the notation, since the above integral mentions the angle, whereas the integral $\int_C f(z)\,dz$ mentions only the position - and hence is resistant to reparameterization.
To be more precise, if $g:\mathbb C\rightarrow \mathbb C$ is thought of as a field of forces at each point, then
$$\operatorname{Re}\left(\int_C\overline{g(z)}\,dz\right)$$
is the energy gained by a point along the curve $C$ in the field $g$. If we set $g(z)=\overline{f(z)}$ to relate this case to the given one, we would see that $g(z)=z$ on the circle $C$, so the force would always be perpendicular to the motion - hence the real part of the given integral is $0$. However, if we want the imaginary part of this integral, we would see that that's the real part of $\int_C \overline{ig(z)}\,dz$ and we would note that the field $ig(z)$ is circulating counterclockwise - hence would always push the point along its direction of motion, thus leading to an increase in energy, the same way that dropping a ball in a constant field of gravity leads to an increase of energy as it falls in the same direction as the force acting on it. Note also that $\int_{|z|=1}$ is not such a good notation, because the physical intuition makes it clear that it matters a lot which way we go (since going counterclockwise, the force helps us along, and going clockwise, it hinders us) and how many times we go around.
We can also just look at this integral formally; we have
$$\int_C \frac{dz}z$$
and if we substitute $z=e^{i\theta}$ which implies $dz=ie^{i\theta}\,d\theta$ we get
$$\int_0^{2\pi}\frac{ie^{i\theta}}{e^{i\theta}}=\int_{0}^{2\pi}i\,d\theta = 2\pi i$$
where we see that the direction in which $z$ is moving cancels, by multiplication, with the value of the function at that $z$. Note that this converts line integrals into the integral of a function from $\mathbb R$ and if we applied this process to $z^{-2}$, we would get the first integral in this post as a result - and, physically, the field given by $\overline{z^{-2}}$ would constantly change its angle to the unit circle over a rotation, hence would not create any energy.
The reason for this is largely the same as why one has $d\theta$ in the real case: the value $f(z)\,dz$ means that the amount of value this term contributes to the integral is the change in $z$ over a small interval times $f(z)$ - that's essentially what Riemann integration formalizes. For complex numbers, the change in $z$ is likely not real, but the value $f(z)$ is sort a hypothetical saying "if $z$ increased by $1$ and the value of the function stayed the same, how much would that segment contribute to the integral?" and if $z$ is moving in some other direction, we have to compensate for that. Essentially*, you can think of the term $f(z)\,dz$ as defining, at each point $z$, a linear function $\mathbb C\rightarrow\mathbb C$ taking "small changes in $z$" to "small changes in the integral" and the term $\frac{dz}z$ just happens to define the function that takes the clockwise tangent to the unit circle to $i$ at all points on the unit circle.

Perhaps another useful example to consider is integrating the constant $1$ around the unit circle; obviously, if we set $f(z)=1$ then $\int_C f(e^{i\theta})\,d\theta = 2\pi$, which reflects that the average value of $f$ is $1$ along the unit circle. Since the term $1\,dz$ represents that a change in $z$ is reflected exactly as a change in the integrated value, the integral
$$\int_C 1\,dz$$ 
represents summing up all of the changes in $z$ over a circle - but since the changes in $z$ are not weighted in any way, this integral is just equal to the total change in $z$ over the curve $C$ - which is $0$ for any closed loop.

One last note which is maybe less explanatory than the others, but is a nice idea that forces ones intuition about everything else into place: really, an integral $\int f(z)\,dz$ is trying to find an antiderivative $F$ of $f$. So, we're essentially stuck only with the rule that $F(z)=f(z)$. This tells us why we have to define the integral the way we did: $f(z)$ is literally the ratio $\frac{F(z+h)-F(z)}{h}$ meaning that $F(z+h)\approx F(z) + h f(z)$, where $h$ is some small change - so this term $h f(z)$ tells us how quickly the antiderivative changes when we move with a velocity of $h$.
If $f$ is defined on a disk (or any simply connected open set), then we can even define an anti-derivative $F$ by choosing a base point $x_0$ and setting $F(z)$ to be the line integral of $f$ along any path from $x_0$ to $z$ - where the choice of path provably will not matter. More strongly, if every integral along a loop works out to zero, this definition gives an antiderivative (...which is a named theorem, but I forget the name). When the integrals over loops fail to come out to $0$, we know that there's some problem in defining an anti-derivative. For $f(z)=1/z$ the issue is that we would like $F(z)=\log(z)$, but the logarithm isn't well defined because $e^{0}=e^{2\pi i}$ - so exponentiation is not invertible. The integral $\int\frac{dz}z$ then can be thought of as considering what happens if we look at the curve $e^{i\theta}$ and try to find a function $h(\theta)$ so that $e^{i\theta}=e^{h(\theta)}$. If we insist that $h$ be continuous, we find out that $h(2\pi) - h(0)$ must be $2\pi i$ - which tells us exactly how $1/z$ fails to have an antiderivative and how $e^z$ fails to have an inverse.

*This is exactly what a differential form is defined to be, in the somewhat trivial case where we are studying $\mathbb C$. We just don't think about it too much because the only linear functions $\mathbb C\rightarrow\mathbb C$ are those that just multiply their input by some constant.
