$\left|\sum_{n=1}^{\infty} \frac{1}{\left(\frac{1}{2}-z_n\right)^2}\right|= \sum_{n=1}^{\infty} \frac{1}{\left|\frac{1}{2}-z_n\right|^2}$

Question

If $z_n\in \mathbb{C} $ and $0<\Re(z_n)<1, \ \forall n\geq 1 $

Given, $\left|\sum_{n=1}^{\infty} \frac{1}{\left(\frac{1}{2}-z_n\right)^2}\right|= \sum_{n=1}^{\infty} \frac{1}{\left|\frac{1}{2}-z_n\right|^2}$ then find the condition on $z_n$ so that the above equality holds.

Attempt

Write, $z_n=a_n+ib_n , 0<a_n<1$ $\Rightarrow|\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-z_n)^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{|\frac{1}{2}-z_n|^2}]^2$

$\Rightarrow |\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n-ib_n)^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$ $\Rightarrow |\sum_{n=1}^{\infty} \frac{(\frac{1}{2}-a_n+ib_n)^2}{[(\frac{1}{2}-a_n)^2+b_n^2]^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$ $\Rightarrow |\sum_{n=1}^{\infty} \frac{((\frac{1}{2}-a_n)^2-b_n^2)+i2b_n(\frac{1}{2}-a_n)}{[(\frac{1}{2}-a_n)^2+b_n^2]^2}|^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$

$\Rightarrow [\sum_{n=1}^\infty\frac{((\frac{1}{2}-a_n)^2-b_n^2)}{ [ (\frac{1}{2}-a_n)^2+b_n^2]^2 }]^2 + [\sum_{n=1}^\infty\frac{2b_n(\frac{1}{2}-a_n)}{ [ (\frac{1}{2}-a_n)^2+b_n^2]^2 }]^2= [\sum_{n=1}^{\infty} \frac{1}{(\frac{1}{2}-a_n)^2+b_n^2}]^2$

Answer 1

Let $w_n = \frac{1}{(\frac12-z_n)^2}$. You need that $$ \left|\sum_{n=1}^\infty w_n \right| = \sum_{n-1}^\infty |w_n|$$ First of all, for the right side of this condition to make sense, we need the series $\sum_{n-1}^\infty |w_n|$ to converge. It happens to be a sufficient condition for the series $\sum_{n=1}^\infty w_n $ to converge, so we don't need an additional condition for the convergence of the left sice of the condition.

Generaly we have $$ \left|\sum_{n=1}^\infty w_n \right| \le \sum_{n-1}^\infty |w_n|$$ (the generalized triangle inequality), with the equality happening only if all $w_n$ are proportional to each other with positive proportionaility coefficientsts, i.e. $$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N: w_n = \lambda_n\omega$$ Going back to $z_n$ this condition can be rewritten as $$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N: \frac{1}{(\frac12- z_n)^2} = \lambda_n\omega$$ $$ \exists \omega\in\mathbb C\, \exists\lambda_n \in \mathbb R_+\, \forall n\in\mathbb N : z_n -\frac12 = \pm\left(\frac{1}{\lambda_n\omega}\right)^\frac12$$ $$ \exists \zeta \in\mathbb C\, \exists\mu_n \in \mathbb R\setminus\{0\} \forall n\in\mathbb N : z_n = \frac12 + \mu_n\zeta$$

Let us assume that ${\rm Re}\, \zeta \neq 0$. From the condition $0 < {\rm Re}\,z_n < 1$ we get then $$ 0 < \frac12 + \mu_n {\rm Re}\,\zeta < 1 $$ $$ -\frac{1}{|{\rm Re}\,\zeta|} < \mu_n < \frac{1}{|{\rm Re}\,\zeta|}$$ We have then $$ |z_n-\frac12| = |\mu_n\zeta| < \frac{|\zeta|}{|{\rm Re}\,\zeta|}$$ $$ |w_n| = \frac{1}{|z_n-\frac12|^2} = \frac{1}{|\mu_n\zeta|^2}> \frac{|{\rm Re}\,\zeta|^2}{|\zeta|^2}$$ For such $|w_n|$ the series $\sum_{n=1}^\infty |w_n|$ would not be convergent. Therefore we need ${\rm Re}\, \zeta = 0$. Without a loss of generality we can choose $\zeta = i$, and we have the conditions $$ z_n = \frac12 + \mu_n i \qquad \text{for some } \mu_n\in\mathbb R\setminus\{0\}$$ $$ \text{series } \sum_{n-1}^\infty |w_n| = \sum_{n-1}^\infty \frac{1}{\mu_n^2} \text{ is convergent}$$