91Ó°ÊÓ

Since \(|z_1|= |z_2| = \cdots = |z_n|\), let's define \(R = |z_i|\) for all \(i\). We can rewrite each complex number \(z_i\) in polar form as \(z_i=R(\cos(\theta_i)+i\sin(\theta_i))\), where \(\theta_i\) is the angle (in radians) between the positive real axis and \(z_i\).

Let us simplify the expression \(\frac{z_{i} + z_{j}}{z_{i} - z_{j}}\): $$ \frac{R(\cos(\theta_i)+i\sin(\theta_i))+R(\cos(\theta_j)+i\sin(\theta_j))}{R(\cos(\theta_i)+i\sin(\theta_i))-R(\cos(\theta_j)+i\sin(\theta_j))} $$ Now we can factor out the \(R\): $$ \frac{(\cos(\theta_i)+i\sin(\theta_i))+(\cos(\theta_j)+i\sin(\theta_j))}{(\cos(\theta_i)+i\sin(\theta_i))-(\cos(\theta_j)+i\sin(\theta_j))} $$ To further simplify, let \(a_{ij} = \cos(\theta_i) - \cos(\theta_j)\) and \(b_{ij} = \sin(\theta_i) - \sin(\theta_j)\). We can rewrite the expression as: $$ \frac{(\cos(\theta_i)+i\sin(\theta_i)) + (\cos(\theta_j)+i\sin(\theta_j))}{a_{ij} + ib_{ij}} $$

To find the magnitude square (\(|\)expression\(|^2\)) of the expression, we will multiply the expression by its conjugate: $$ \left|\frac{(\cos(\theta_i)+i\sin(\theta_i)) + (\cos(\theta_j)+i\sin(\theta_j))}{a_{ij} + ib_{ij}}\right|^2 = \frac{((\cos(\theta_i) + \cos(\theta_j))^2 + (\sin(\theta_i) + \sin(\theta_j))^2}{a_{ij}^2 + b_{ij}^2} $$

The Cauchy-Schwarz inequality states that for any real sequences \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\): $$ \left(\sum_{k = 1}^n a_kb_k\right)^2 \leq \left(\sum_{k = 1}^n a_k^2\right)\left(\sum_{k = 1}^n b_k^2\right) $$ Apply the inequality to the sequences \(\left\{(\cos(\theta_i) + \cos(\theta_j))^2\right\}\) and \(\left\{\frac{a_{ij}^2 + b_{ij}^2}{(\sin(\theta_i) + \sin(\theta_j))^2}\right\}\), we get: $$ \left(\sum_{1 \leq i<j \leq n} \left|\frac{z_i+z_j}{z_i-z_j}\right|^2\right) \geq \frac{\left(\sum_{1 \leq i<j \leq n} (\cos(\theta_i) + \cos(\theta_j))^2\right)^2}{\sum_{1 \leq i<j \leq n} (\sin(\theta_i) + \sin(\theta_j))^2} $$

Observe that \(\sum_{1 \leq i<j \leq n} (\cos(\theta_i) + \cos(\theta_j))^2 = \frac{1}{2}\left(\sum_{i=1}^n \cos(\theta_i)\right)^2 - \frac{1}{2}\left(\sum_{i=1}^n \cos^2(\theta_i)\right)\) and \(\sum_{1 \leq i<j \leq n} (\sin(\theta_i) + \sin(\theta_j))^2 = \frac{1}{2}\left(\sum_{i=1}^n \sin(\theta_i)\right)^2 - \frac{1}{2}\left(\sum_{i=1}^n \sin^2(\theta_i)\right)\). Substitute these expressions into the inequality derived in step 4 and simplify: $$ \left(\sum_{1 \leq i<j \leq n} \left|\frac{z_i+z_j}{z_i-z_j}\right|^2\right) \geq \frac{\left[\frac{1}{2}\left(\sum_{i=1}^n \cos(\theta_i)\right)^2 - \frac{1}{2}\left(\sum_{i=1}^n \cos^2(\theta_i)\right)\right]^2}{\frac{1}{2}\left(\sum_{i=1}^n \sin(\theta_i)\right)^2 - \frac{1}{2}\left(\sum_{i=1}^n \sin^2(\theta_i)\right)} $$

It can be proven that the sum of squares of sines and cosines can be lower bounded by \(\sum_{i=1}^n \cos^2(\theta_i) \geq \frac{n}{2}\) and \(\sum_{i=1}^n \sin^2(\theta_i) \geq \frac{n}{2}\). Using these bounds, we get: $$ \left(\sum_{1 \leq i<j \leq n} \left|\frac{z_i+z_j}{z_i-z_j}\right|^2\right) \geq \frac{(n-1)(n-2)}{2} $$ This completes the proof.