Let \(f\) be one-to-one and analytic on a neighborhood of the unit circle \(S^1\subset\mathbb C\). Relate \[ \int_{S^1}\overline{f(z)}\,f'(z)\,dz \] to the area of the region enclosed by \(f(S^1)\).
Solution
Let \(A\) denote the ordinary (unsigned) area of the bounded component of \(\mathbb C\setminus f(S^1)\). Then \[ \left|\int_{S^1}\overline{f(z)}\,f'(z)\,dz\right|=2A. \] More precisely, if \(f(S^1)\) is given the orientation induced from the counterclockwise orientation of \(S^1\), then \[ \int_{S^1}\overline{f(z)}\,f'(z)\,dz =2iA_{\mathrm{sgn}}, \] where \(A_{\mathrm{sgn}}=A\) for counterclockwise orientation and \(A_{\mathrm{sgn}}=-A\) for clockwise orientation.
1. The restriction of \(f\) to \(S^1\) parametrizes a \(C^1\) Jordan curve.
Because \(f\) is one-to-one on a neighborhood of \(S^1\), its restriction to \(S^1\) is injective. Since \(S^1\) is compact and \(\mathbb C\) is Hausdorff, \[ f|_{S^1}:S^1\longrightarrow f(S^1) \] is a homeomorphism. Hence \(f(S^1)\) is a Jordan curve.
Moreover \(f'(z)\neq0\) on \(S^1\). Indeed, if \(f'(z_0)=0\), then the local power-series expansion at \(z_0\) has the form \[ f(z)-f(z_0)=(z-z_0)^m h(z), \qquad m\ge2,\quad h(z_0)\neq0, \] so sufficiently near \(z_0\) the map has local multiplicity \(m\) and cannot be one-to-one. Thus the parametrization \[ \gamma(t)=f(e^{it}),\qquad 0\le t\le2\pi, \] is regular and \(C^1\).
2. The given integral is the line integral\[ \int_{f(S^1)}\overline w\,dw \]with the orientation induced by \(f\).
Along the parametrization \(w=f(z)\) one has \(dw=f'(z)\,dz\). Therefore \[ \int_{S^1}\overline{f(z)}\,f'(z)\,dz =\int_{f(S^1)}\overline w\,dw. \]
3. If \(w=x+iy\), then\[ \int_{f(S^1)}\overline w\,dw =i\int_{f(S^1)}(x\,dy-y\,dx). \]
Since \[ \overline w\,dw =(x-iy)(dx+i\,dy) =x\,dx+y\,dy+i(x\,dy-y\,dx), \] the real part integrates to zero around the closed curve because \[ x\,dx+y\,dy=\frac12\,d(x^2+y^2). \] Hence only the displayed imaginary term remains.
4. Green’s theorem gives\[ \int_{f(S^1)}(x\,dy-y\,dx)=2A_{\mathrm{sgn}}. \]
Let \(\Omega\) be the bounded component enclosed by the Jordan curve \(f(S^1)\). For the counterclockwise boundary orientation, Green’s theorem applied to the \(1\)-form \(x\,dy-y\,dx\) gives \[ \int_{\partial\Omega}(x\,dy-y\,dx) =\iint_\Omega \left(\frac{\partial x}{\partial x} -\frac{\partial(-y)}{\partial y}\right)\,dx\,dy =2\iint_\Omega dx\,dy =2A. \] Reversing the orientation changes the line integral by a sign. Thus for the orientation induced by \(f\) the value is \(2A_{\mathrm{sgn}}\).
5. Therefore\[ \int_{S^1}\overline{f(z)}\,f'(z)\,dz=2iA_{\mathrm{sgn}}, \qquad A=\frac12\left|\int_{S^1}\overline{f(z)}\,f'(z)\,dz\right|. \]
Combine <1>2–<1>4. The absolute-value formula removes the orientation ambiguity. This ambiguity is genuine under the stated hypotheses: \(f(z)=z\) induces the counterclockwise orientation, whereas \(f(z)=1/z\), which is holomorphic and one-to-one on an annular neighborhood of \(S^1\), induces the clockwise orientation.