Let $f : \mathbb C → \mathbb C$ be a function of class $C^1$
(not necessarily holomorphic); write $f = u + iv$.
Let $Ω ⊂ C$ be a domain with boundary $bΩ = C$, where $C$ is a simple, closed, piecewise differentiable curve.
Define the planar vector fields $F_1 = ui − vj$,
$F_2 = vi + uj$.
(a) Show that $\int_C{f(z)dz} = \int_C{F1 · dr} +i \int_C{F2 · dr} $.
(b) Use Green’s theorem to show that $\int_C{f(z)dz} = 2i \int \int_Ω {\frac{∂f} {∂\bar{z}}} dxdy$.
(c) What can we conclude if $f$ is $C^1$ and complex differentiable?
My attempts:
a) writing $f(z) = u(x, y) + iv(x, y)$ and $dz = dx+idy$, and $r(t) = (x(t), y(t))$, we have:
$\int_C f(z) dz = \int_a^b({u(x(t), y(t))}\frac{dx(t)}{dt} − v(x(t), y(t))\frac{dy(t)}{dt}) dt + i\int_a^b({u(x(t), y(t))} \frac{dy(t)}{dt} + v(x(t), y(t))\frac{dx(t)}{dt}) dt = \int_C F_1.dr + i\int_C F_2.dr$
b) I only know Green's theorem for real-valued functions but don't know how to apply it here. Please could you show me how to solve this part?
c) I know that if $f$ is comolex differentiable then it is differentiable in the real sense as a map from $\mathbb R^2$ to $\mathbb R^2$ and its differential is $f'(z)$. I also know that $\frac{\partial f} {\partial {\bar{z}}}$ =
$1/2(\frac{\partial f}{\partial x}+ i \frac{\partial f}{\partial y})$, but how do I proceed from here and what does this tell me? Please any help.
Best Answer
b) Apply Green's theorem to the result you got from a). They are real integrals.
c) Complex differentiable means $\partial f/\partial\bar z=0$.