Prove the integral of $Re(z)$ around a simple closed curve is imaginary

Question

If $θ$ is any simple closed curve in $\mathbb{C}$, prove that $\int_{θ}Re(z)dz$ is pure imaginary

I have shown if $θ$ is the unit circle then the answer is $\frac{i}{2}$ but I'm struggling with a general proof.

I can prove it for any circle $C(a;R)$, is it possible maybe we can represent any simple closed curve as a union of circles? That's all I can think of.

Any help would be hugely appreciated.

Answer 1

Well, let us note $$\gamma : t\in [0,1] \to \gamma(t) \in \theta$$ a parametrization of $\theta.$ Then by definition we have $$\int_{θ}\Re(z)dz =\int_0^1 \Re(\gamma(t))\gamma'(t)dt = \int_0^1 \Re(\gamma(t))\Re(\gamma'(t))dt+i\int_0^1 \Re(\gamma(t))\Im(\gamma'(t))dt.$$ But by definition $$\Re(\gamma'(t))=\Re(\gamma(t))'.$$ Hence if we note $f(t) = \Re(\gamma(t))$ your thesis is to prove that $$\int_0^1 f(t)f'(t) dt = 0.$$ Let us do an integration by part $$\int_0^1 f(t)f'(t) dt=[f(t)f(t)]_0^1-\int_0^1 f(t)f'(t) dt.$$ Hence $$2 \int_0^1 f(t)f'(t) dt = [f(t)f(t)]_0^1 =0$$ since $f(1) = f(0)$ because the curve is closed.

Answer 2

Notice that $\Re(z)dz + \Re(\bar z)d\bar z = \frac 12(z dz + \bar z dz + z d\bar z + \bar z d\bar z) = \frac 14 d(z^2 + 2z\bar z + \bar z^2) = d(\Re(z)^2)$ is a closed $1$-form.

So if you integrate over a closed curve, $\Re(\int \Re(z)dz) = \frac 12 \int (\Re(z)dz + \Re(\bar z)d\bar z) = 0$