Applied Mathematics/Complex Integration

Complex integration
On the piecewise smooth curve $$C: z = z(t)$$ $$(a\leqq t \leqq b)$$, suppose the function f(z) is continuous. Then we obtain the equation below.
 * $$ \int_C f(z)dz = \int_a^b f\{z(t)\}\ \frac{dz(t)}{dt}dt$$

where $$f(z)$$ is the complex function, and $$z$$ is the complex variable.

Proof
Let
 * $$f(z)=u(x,y)+iv(x,y)$$
 * $$dz=dx+idy $$

Then
 * $$\int_C f(z)dz$$
 * $$=\int_C(u+iv)(dx+idy)$$
 * $$= \left( \int_C udx - \int_C vdy \right) + i\left( \int_C vdx - \int_C udy  \right)$$

The right side of the equation is the real integral, therefore, according to calculus, the relationship below can be applied.
 * $$\int_{x_1}^{x_2}f(x)dx=\int_{t_1}^{t_2}f(x)\frac{dx}{dt}dt$$

Hence
 * $$\int_C f(z)dz$$
 * $$= \left( \int_a^b u \frac{dx}{dt} dt - \int_a^b v \frac{dy}{dt}dt \right) + i \left( \int_a^b v \frac{dx}{dt}dt - \int_a^b u \frac{dy}{dt}dt \right)$$
 * $$= \left( \int_a^b u x'(t)dt - \int_a^b v y'(t)dt \right) + i \left( \int_a^b v x'(t)dt - \int_a^b u y'(t)dt \right)$$
 * $$=(u+iv) \left( x'(t)+ iy'(t) \right)dt$$
 * $$=\int_a^b f\{z(t)\}\ z'(t)dt$$

This completes the proof.