General way to find out whether a curve is positively oriented I have a general question, that deals with the question how I am able to find out whether a particular curve(in $\mathbb{C}$) is positively oriented?  Take e.g. $ y(t)=a+re^{it}$. Obviously this one is positively oriented, but is there a fast general method to proof this? 
 A: The question you are asking is how to prove that the turning number of a simple closed curve is equal to one.  This is the same as saying that the winding number of the curve is one around any point lying inside the curve.
There are several ways to do this, and none of them is particularly fast.  Here are some possibilities:


*

*One basic possibility is to make a graph of $\arg(\gamma'(t))$.  If the argument increases by $2\pi$ as you go once around the curve (ignoring any jumps from $\pi$ to $-\pi$ on your graph), then the curve is positively oriented.  Similarly, if $p$ is any point lying inside the curve, you could make a graph of $\arg(\gamma(t) - p)$ and do the same thing.  Of course, both of these involve drawing a graph, which you may prefer not to do.
On a related note, given a closed curve $\gamma\colon [a,b]\to\mathbb{C}$, if you can find a continuous function $\varphi\colon [a,b]\to \mathbb{C}$ so that $\gamma'(t) = e^{\varphi(t)}$, then the curve is positively oriented if and only if $\varphi(b)-\varphi(a)=2\pi i$, and negatively oriented if and only if $\varphi(b)-\varphi(a)=-2\pi i$.  This is certainly the fastest way of dealing with the example you gave.
Similarly, if $p$ is any point inside the curve, you could find a function $\varphi(t)$ so that $\gamma(t) - p =e^{\varphi(t)}$, and perform the same test.

*Assuming the curve $\gamma\colon [a,b]\to \mathbb{C}$ is twice differentiable (with the derivatives at $a$ and $b$ matching up) and has no critical points, you can integrate the curvature of the curve along its length:
$$
\frac{1}{2\pi}\int_a^b \frac{d}{dt}[\arg(\gamma'(t))]\,dt \;=\; \frac{1}{2\pi}\int_a^b \frac{\mathrm{Re}[i\,\gamma'(t)\,\gamma''(t)]}{|\gamma'(t)|^2} dt,
$$
This integral will come out to $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.

*If $p$ is a point lying inside the curve, you can use an integral to compute the winding number around $p$.  This requires that the curve $\gamma\colon [a,b]\to \mathbb{C}$ be differentiable:
$$
\frac{1}{2\pi}\int_a^b \frac{d}{dt}[\arg(\gamma(t)-p)]\,dt \;=\; \frac{1}{2\pi}\int_a^b \frac{\mathrm{Re}[i\,(\gamma(t)-p)\,\gamma'(t)]}{|\gamma(t)-p|^2} dt,
$$
Again, this integral will come out to $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.  By the way, if this integral comes out to $0$ it means that the point $p$ you chose does not lie inside the curve.  Indeed, you can use this integral to test whether a given point lies inside a given closed curve.

*You can compute the degree of $\gamma'$ as a sum of local degrees.  Specifically, let $\theta$ be any fixed angle, and let $t_1,\ldots,t_n$ be the values of $t$ for which $\arg(\gamma'(t)) = \theta$.  For each $k\in\{1,\ldots,n\}$, let
$$
d_k \;=\; \begin{cases}+1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma''(t_k))>0 \\[6pt] -1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma''(t_k))<0 \end{cases}
$$
(If the imaginary part every comes out to zero, you should probably just choose a different value of $\theta$.)  Then the sum
$$
d_1 + \cdots + d_n
$$
will be $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.
This test is easier to apply than it looks.  For many curves, there will only be one value of $t$ for which $\arg(\gamma'(t)) = \theta$, so you only need to check whether $\mathrm{Im}(e^{-i\theta}\gamma'(t))$ is positive or negative for this one value of $t$.

*Similarly, if $p$ is a point lying inside the curve, you can compute the winding number of $\gamma$ around $p$ using a sum of local degrees. Specifically, let $\theta$ be any fixed angle, and let $t_1,\ldots,t_n$ be the values of $t$ for which $\arg(\gamma(t)-p) = \theta$.  For each $k\in\{1,\ldots,n\}$, let
$$
d_k \;=\; \begin{cases}+1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma'(t_k))>0 \\[6pt] -1 & \text{if }\mathrm{Im}(e^{-i\theta}\gamma'(t_k))<0 \end{cases}
$$
(If the imaginary part every comes out to zero, you should probably just choose a different value of $\theta$.)  Then the sum
$$
d_1 + \cdots + d_n
$$
will be $+1$ if the curve is positively oriented, and $-1$ if the curve is negatively oriented.
Again, this test is easier to apply than it looks, since there is often just one value.  As with test #3, if the sum comes out to $0$ it means that the point $p$ does not in fact lie in the interior of the curve.

*Given a differentiable curve $\gamma\colon [a,b] \to \mathbb{C}$, find a partition of $[a,b]$ into intervals $I_1,\ldots,I_n$ with the following property: for each $k$, the image of $\gamma'$ on the closure of $I_k$ lies in one of the following half-planes:
$$
H_0 = \{\mathrm{Re}(z) > 0\},\quad H_1 = \{\mathrm{Im}(z) > 0\},\quad H_2 = \{\mathrm{Re}(z) < 0\},\quad H_3 = \{\mathrm{Im}(z) < 0\}.
$$
Note that these half-planes overlap, so you don't need to be particularly careful in choosing your partition.  For each $k$, let $m_k\in \{0,1,2,3\}$ be the number of the half-plane containing $I_k$.  For each transition $(m_k,m_{k+1})$, label it $+1$ if $m_{k+1} \equiv m_k +1 (\mathrm{mod}\;4)$, and $-1$ if $m_{k+1} \equiv m_k - 1 (\mathrm{mod}\;4)$.  Then your curve is positively oriented if there are more $+1$ transitions than $-1$'s and negatively oriented if there are more $-1$ transitions than $+1$'s.
A similar approach can be used on $\gamma(t)-p$ instead to compute the winding number, where $p$ is a point inside of the curve.
A: Orientation only really applies to simple, closed curves. 
For a general curve, you might find the winding number to be useful.
See: http://mathworld.wolfram.com/ContourWindingNumber.html
