Complex integration

On the piecewise smooth curve ${\displaystyle C:z=z(t)}$ ${\displaystyle (a\leqq t\leqq b)}$, suppose the function f(z) is continuous. Then we obtain the equation below.

${\displaystyle \int _{C}f(z)dz=\int _{a}^{b}f\{z(t)\}\ {\frac {dz(t)}{dt}}dt}$

where ${\displaystyle f(z)}$ is the complex function, and ${\displaystyle z}$ is the complex variable.

Proof

Let

${\displaystyle f(z)=u(x,y)+iv(x,y)}$

${\displaystyle dz=dx+idy}$

Then

${\displaystyle \int _{C}f(z)dz}$

${\displaystyle =\int _{C}(u+iv)(dx+idy)}$

${\displaystyle =\left(\int _{C}udx-\int _{C}vdy\right)+i\left(\int _{C}vdx-\int _{C}udy\right)}$

The right side of the equation is the real integral, therefore, according to calculus, the relationship below can be applied.

${\displaystyle \int _{x_{1}}^{x_{2}}f(x)dx=\int _{t_{1}}^{t_{2}}f(x){\frac {dx}{dt}}dt}$

Hence

${\displaystyle \int _{C}f(z)dz}$

${\displaystyle =\left(\int _{a}^{b}u{\frac {dx}{dt}}dt-\int _{a}^{b}v{\frac {dy}{dt}}dt\right)+i\left(\int _{a}^{b}v{\frac {dx}{dt}}dt-\int _{a}^{b}u{\frac {dy}{dt}}dt\right)}$

${\displaystyle =\left(\int _{a}^{b}ux'(t)dt-\int _{a}^{b}vy'(t)dt\right)+i\left(\int _{a}^{b}vx'(t)dt-\int _{a}^{b}uy'(t)dt\right)}$

${\displaystyle =(u+iv)\left(x'(t)+iy'(t)\right)dt}$

${\displaystyle =\int _{a}^{b}f\{z(t)\}\ z'(t)dt}$

This completes the proof.