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.

$\underset{C}{\int }f(z)dz=\underset{a}{\overset{b}{\int }}f\{z(t)\}\frac{dz(t)}{dt}dt$

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

$\underset{C}{\int }f(z)dz$

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

$=(\underset{C}{\int }udx-\underset{C}{\int }vdy)+i(\underset{C}{\int }vdx-\underset{C}{\int }udy)$

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

$\underset{x_1}{\overset{x_2}{\int }}f(x)dx=\underset{t_1}{\overset{t_2}{\int }}f(x)\frac{dx}{dt}dt$

Hence

$\underset{C}{\int }f(z)dz$

$=(\underset{a}{\overset{b}{\int }}u\frac{dx}{dt}dt-\underset{a}{\overset{b}{\int }}v\frac{dy}{dt}dt)+i(\underset{a}{\overset{b}{\int }}v\frac{dx}{dt}dt-\underset{a}{\overset{b}{\int }}u\frac{dy}{dt}dt)$

$=(\underset{a}{\overset{b}{\int }}u{x}^{\prime }(t)dt-\underset{a}{\overset{b}{\int }}v{y}^{\prime }(t)dt)+i(\underset{a}{\overset{b}{\int }}v{x}^{\prime }(t)dt-\underset{a}{\overset{b}{\int }}u{y}^{\prime }(t)dt)$

$=(u+iv)(x^{\prime }(t)+i{y}^{\prime }(t))dt$

$=\underset{a}{\overset{b}{\int }}f\{z(t)\}{z}^{\prime }(t)dt$

This completes the proof. Template:BookCat