Let \(f\) be holomorphic on a neighborhood of the closed unit disk, and let \(L\) be the length of the image of the unit circle under \(f\). Show that \[ L\ge2\pi|f'(0)|. \]
Solution
1. The length \(L\) is\[ L=\int_0^{2\pi}\abs{f'(e^{it})}\,dt. \]
Parametrize the unit circle by \[ \gamma(t)=e^{it}, \qquad 0\le t\le2\pi. \] The image curve is \(f\circ\gamma\), whose derivative is \[ (f\circ\gamma)'(t) = ie^{it}f'(e^{it}). \] Since \(\abs{ie^{it}}=1\), its speed is \(\abs{f'(e^{it})}\). The arc-length formula therefore gives the displayed identity.
2. One has\[ 2\pi f'(0) = \int_0^{2\pi} f'(e^{it})\,dt. \]
Since \(f\) is holomorphic on a neighborhood of the closed unit disk, \(f'\) is holomorphic there as well. Cauchy’s integral formula applied to \(f'\) at \(0\) gives \[ f'(0) = \frac{1}{2\pi i} \int_{\abs{z}=1}\frac{f'(z)}{z}\,dz. \] With \(z=e^{it}\) and \(dz=ie^{it}\,dt\), this becomes \[ f'(0) = \frac{1}{2\pi} \int_0^{2\pi} f'(e^{it})\,dt, \] which is the claimed identity.
3. The required estimate is\[ \boxed{L\ge2\pi\abs{f'(0)}}. \]
By step 2 and the triangle inequality, \[ 2\pi\abs{f'(0)} = \abs{\int_0^{2\pi}f'(e^{it})\,dt} \le \int_0^{2\pi}\abs{f'(e^{it})}\,dt. \] The right-hand side is \(L\) by step 1.
4. Q.E.D.
Step 3 is the required inequality.