# 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

$\int _{C}f(z)dz$ $=\left(\int _{a}^{b}u{\frac {dx}{dt}}dt-\int _{a}^{b}v{\frac {dy}{dt}}dt\right)+i\left(\int _{a}^{b}v{\frac {dx}{dt}}dt-\int _{a}^{b}u{\frac {dy}{dt}}dt\right)$ $=\left(\int _{a}^{b}ux'(t)dt-\int _{a}^{b}vy'(t)dt\right)+i\left(\int _{a}^{b}vx'(t)dt-\int _{a}^{b}uy'(t)dt\right)$ $=(u+iv)\left(x'(t)+iy'(t)\right)dt$ $=\int _{a}^{b}f\{z(t)\}\ z'(t)dt$ This completes the proof.