Consider the series of complex functions: \[ \sum_{n=1}^\infty \frac{\sin(nz)}{2^n}. \] (a) Determine whether the series converges uniformly on the half-plane \(\{ z \in \mathbb{C} \mid \operatorname{Im}(z) < \ln 2 \}\), or clarify the correct domain of uniform convergence (such as \(\{ \operatorname{Im}(z) \ge -c \}\) or compact subsets \(\{ |\operatorname{Im}(z)| \le Y \}\) / disks \(\{ |z| \le r \}\)). (b) Prove that for every \(r\) with \(0<r<\ln 2\), the series converges uniformly on the closed disk \(\{ z \in \mathbb{C} \mid |z| \le r \}\).
Solution
Write \(z=x+iy\). Since \[ \sin(nz)=\frac{e^{inz}-e^{-inz}}{2i}, \] we have \[ |\sin(nz)|\le \frac{e^{-ny}+e^{ny}}2\le e^{n|y|}. \] Hence \[ \left|\frac{\sin(nz)}{2^n}\right| \le \left(\frac{e^{|\Im z|}}2\right)^n. \]
Therefore, on every closed substrip \[ |\Im z|\le Y<\log2, \] the series converges uniformly by the Weierstrass \(M\)-test, with majorant \((e^Y/2)^n\).
Conversely, if \(|\Im z|>\log2\), then one of the two exponentials in \(\sin(nz)\) dominates and \[ \frac{|\sin(nz)|}{2^n}\not\longrightarrow0. \] For example, at \(z=-iy\) with \(y>\log2\), \[ \frac{|\sin(nz)|}{2^n}=\frac{\sinh(ny)}{2^n}\sim\frac12\left(\frac{e^y}{2}\right)^n. \] Thus the pointwise convergence region is exactly \[ |\Im z|<\log2. \] In particular, the series is not uniformly convergent on the half-plane \(\Im z<\log2\), since that half-plane contains points with imaginary part below \(-\log2\) where the series diverges.
Finally, if \(0<r<\log2\) and \(|z|\le r\), then \(|\Im z|\le r\), so \[ \left|\frac{\sin(nz)}{2^n}\right| \le\left(\frac{e^r}{2}\right)^n, \] and the \(M\)-test gives uniform convergence on the closed disk \(|z|\le r\).