Using Green's theorem for complex valued functions

Question

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.

Answer 1

b) Apply Green's theorem to the result you got from a). They are real integrals.

c) Complex differentiable means $\partial f/\partial\bar z=0$.

Answer 2

we have $\oint u\ dx - v\ dy + i\oint v\ dx + u \ dy$

These are real-valued functions $(x,y)$ are real, and $u,v$ take real variables to real variables.

Green's theorem says that if you have a path integral over a closed curve, you can turn this into an integral over the enclosed surface. Applying Green's to each path integral separately.

$\iint -\frac {\partial u}{\partial y} - \frac {\partial v}{\partial x} \ dy\ dx+i\iint -\frac {\partial v}{\partial y} + \frac {\partial u}{\partial x} \ dy\ dx$

We can unify these.

$\iint (-\frac {\partial u}{\partial y} - \frac {\partial v}{\partial x}) + i(-\frac {\partial v}{\partial y} + \frac {\partial u}{\partial x}) \ dy\ dx$

If $f(z)$ is analytic, what do the Cauchy-Reimann equations tell us?

$\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y}\\ \frac {\partial u}{\partial y} = -\frac {\partial v}{\partial x}$

Which says that these integrals evaluate to $0.$