expressing contour integral in different form

Question

Hi I have a short question regarding contour integration: Given that $f(z)$ is a continuous function over a rectifiable contour $z = x + iy$. If $f(z) = u(x,y) + iv(x,y)$, why does it follow that the contour integral can be expressed as $$\int_{C}f(z)dz = \int_{C}(udx-vdy) + i \int_{C}(vdx +udy)$$ where on the right-hand side, $C$ is a rectifiable curve in the $xy$-plane.

Answer 1

Hint: $$\int_{C}f(z)dz = \int_{C}[u(x,y)+iv(x,y)](dx+idy)$$

Answer 2

Another way to see it is by expanding the integral

$$ \int_\gamma f(z)dz = \int_a^bf(\gamma(t))\gamma'(t)dt = \int_a^b(u(\gamma_1(t),\gamma_2(t)) + i v(\gamma_1(t),\gamma_2(t)))(\gamma_1(t) + i\gamma_2(t))dt, $$

where $\gamma(t) = (\gamma_1(t),\gamma_2(t)), t\in [a,b],$

and applying the definition of the line integrals

$$ \int_\gamma p dx = \int_a^b p(\gamma(t))\gamma_1'(t)dt $$

and

$$ \int_\gamma q dy = \int_a^b q(\gamma(t))\gamma_2'(t)dt $$

from calculus.