## How do I find the value of the real parameter?

Interpreting this series as a sum of two geometric series, we see that this converges when both \displaystyle \left| -\frac{1}{9} (z + 4 + i\lambda)^2 \right| < 1 \text{ and } \left| -\frac{1}{9} (z + 2i - \lambda)^2 \right| < 1. \tag*{} Rewriting these inequalities yields \displaystyle |z - (-4 - i\lambda)| < 3 \text{ and } |z - (\lambda - 2i)| < 3. \tag*{} Hence, the region of convergence is the intersection of the interiors of two disks both with radius 3 and having centers (-4, -\lambda) and (\lambda, -2). The area of this region is maximized when the distance between these disks centers d is minimized. By the distance formula, d = \sqrt{(\lambda + 4)^2 + (\lambda - 2)^2} = \sqrt{2(\lambda + 1)^2 + 8}. \tag*{} Since we have a quadratic under the radical, the minimum value occurs when \boxed{\lambda = -1}. Moreover, d = \sqrt{8}. It now remains to find the area of the overlapping region between the two open disks both with radius 3 and having centers (-4, 1) and (-1, -2). A plot of the required region can be found below: We state a more general version to the area formula we need. Fact: Given two circles having radii r_1 and r_2 (with r_1 \geq r_2), whose centers are d units from each other such that d < r_1 +r_2 and d > r_1 - r_2 (so that the circles intersect in a manner as in the picture above), the area A of the intersection of the circles equals A = \displaystyle r_1^2 \arccos\Big(\frac{d_1}{r_1}\Big) - d_1 \sqrt{r_1^2 - d_1^2} + r_2^2 \arccos\Big(\frac{d_2}{r_2}\Big) - d_2 \sqrt{r_2^2 - d_2^2}, \tag*{} where d_1 = \displaystyle \frac{r_1^2 - r_2^2 + d^2}{2d} \text{ and } d_2 = d - d_1. \tag*{} An exposition and proof of this result (which I highly recommend looking at) can be found at the following reference: For our question, we have r_1 = r_2 = 3, d = \sqrt{8} = 2\sqrt{2}, and thus d_1 = d_2 = \sqrt{2}. Therefore, the area of the intersection of the circles equals \boxed{\begin{align*} A &= \displaystyle 2 \Big(3^2 \arccos\Big(\frac{\sqrt{2}}{3}\Big) - \sqrt{2} \sqrt{3^2 - (\sqrt{2})^2}\Big)\\ &= 18 \arccos\Big(\frac{\sqrt{2}}{3}\Big) - 2\sqrt{14}. \end{align*}} \tag*{}