Let \(s\in\mathbb R\) and define on \(\mathbb C\setminus(-\infty,0]\) \[ u(re^{i\theta})=r^s\cos(s\theta), \qquad r>0,\ -\pi<\theta<\pi. \] Prove that \(u\) is harmonic.
Solution
Let \[ \Omega\coloneqq\CC\setminus(-\infty,0]. \]
1. The principal logarithm\[ \Log z=\log r+i\theta, \qquad z=re^{i\theta},\ -\pi<\theta<\pi, \]is holomorphic on \(\Omega\).
This is the standard holomorphic branch of the logarithm on the slit plane \(\Omega\).
2. The function\[ F(z)\coloneqq e^{s\Log z} \]is holomorphic on \(\Omega\).
By step 1, \(\Log\) is holomorphic on \(\Omega\). Multiplication by the real constant \(s\) and composition with the entire exponential function preserve holomorphy.
3. For \(z=re^{i\theta}\in\Omega\),\[ \operatorname{Re}F(z)=r^s\cos(s\theta)=u(z). \]
Using the definition of the principal logarithm, \[ \begin{aligned} F(re^{i\theta}) &=e^{s(\log r+i\theta)}\\ &=r^s e^{is\theta}\\ &=r^s\bigl(\cos(s\theta)+i\sin(s\theta)\bigr). \end{aligned} \] Taking real parts gives the identity.
4. The function \(u\) is harmonic on \(\Omega\).
The real part of a holomorphic function is harmonic. By steps <1>2–<1>3, \(u=\operatorname{Re}F\) with \(F\) holomorphic on \(\Omega\). Therefore \[ \boxed{\Delta u=0\text{ on }\Omega}. \]
5. Q.E.D.
Step 4 is the required conclusion.