Applied Mathematics/Complex Integration

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Complex integration[edit]

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[edit]

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.