Describe and sketch the region in the complex plane where \[ \sum_{n=1}^{\infty}\frac1{n^2}\exp\left(\frac{nz}{z-2}\right) \] converges.
Solution
For \(z\neq2\), put \[ w\coloneqq\frac{z}{z-2}. \]
1. The series converges if and only if\[ \Re w\leq0. \]
Its \(n\)th term has absolute value \[ \frac1{n^2}\abs{e^{nw}} = \frac{e^{n\Re w}}{n^2}. \] If \(\Re w<0\), these absolute values are bounded by a convergent geometric series for all sufficiently large \(n\), so the given series converges absolutely. If \(\Re w=0\), the series again converges absolutely because its absolute values are \(1/n^2\). If \(\Re w>0\), then \[ \frac{e^{n\Re w}}{n^2}\longrightarrow\infty, \] so the terms do not tend to zero and the series diverges.
2. If \(z=x+iy\neq2\), then\[ \Re\left(\frac{z}{z-2}\right) = \frac{(x-1)^2+y^2-1}{(x-2)^2+y^2} = \frac{\abs{z-1}^2-1}{\abs{z-2}^2}. \]
Multiplying numerator and denominator by \(\overline z-2\) gives \[ \frac{z}{z-2} = \frac{z(\overline z-2)}{\abs{z-2}^2}. \] The real part of the numerator is \[ \abs z^2-2\Re z = x^2+y^2-2x = (x-1)^2+y^2-1. \] Since \(z\neq2\), the denominator is positive.
3. The region of convergence is\[ \boxed{\{z\in\CC:\abs{z-1}\leq1\}\setminus\{2\}}. \]
By step 2, the inequality in step 1 is equivalent to \[ \abs{z-1}^2-1\leq0, \] hence to \(\abs{z-1}\leq1\). The point \(z=2\) must be removed because the summand is not defined there. Thus the sketch is the closed disk of radius \(1\) centered at \(1\) on the real axis, with the boundary point \(2\) deleted.
4. Q.E.D.
Steps <1>1–<1>3 determine and describe the complete convergence region.