What??
Let me say it another way. Draw a curve in the complex plane that doesn't cross itself, and let it start and end at the same point. In other words it is simple and closed. Also, the curve is smooth, except for a finite number of corners. Along the "smooth" sections, the path function p(t) that defines the curve is differentiable. We will define a contour integral around this curve.
sometimes there aren't enough words for all the things mathematicians want to do. We have a line integral, and a path integral, and now a contour integral, all running along a curve inside some kind of field. Actually the contour integral is similar to the line integral. The complex function f has two input variables and two output variables, somewhat like a force field. But instead of the dot product p′.f, we're going to use the complex product p′×f. Here is the definition.
The contour integral of f around p is the integral of f(p(t))×p′(t). If the curve has corners, compute the integral for each smooth section and add them up.
Reparameterize time by replacing t with u(s). Now p′(t) becomes p′(u(s))×u′(s) by the chain rule. But aha, this is the same as integration by substitution. The answer does not change. We can run along the curve quickly or slowly, stop, even back up and go forward again. the contour integral depends only on the function f and the curve p(t). We saw the same thing when we reparameterized the line integral.
Let f(z) = c, where c is a constant. Integrate c×p′(t) around a closed loop. If p starts and ends at a, the answer is cp(a)-cp(a), or 0.
Next let f(z) = z, and let p(t) = u(t)+v(t)i. This produces the following integral.
∫ uu′ - vv′ + (uv′+vu′)i
The integral becomes (u2-v2)/2 + uvi, or z2/2. when the curve starts and ends at a, we have a2/2-a2/2, or 0. The contour integral is 0 whenever f is a linear function of z.