Actually, $ u + rac{1}{u} = 2 $ implies $ u = 1 $ (since for complex $u$, this only holds if $|u| = 1$ and conjugate symmetry, but equality to 2 forces $u = 1$ in $ \mathbb{C} $ only if real and positive, but let's solve generally).

February 22, 2026

$Actually, $ u + rac{1}{u} = 2 $ implies $ u = 1 $ (since for complex $u$, this only holds if $|u| = 1$ and conjugate symmetry, but equality to 2 forces $u = 1$ in $ \mathbb{C} $ only if real and positive, but let's solve generally).$

rac{z + w}{z - w} = 1 \Rightarrow z + w = z - w \Rightarrow 2w = 0 \Rightarrow w = 0.
But this contradicts \( w
Wait â unless the equation allows limit behavior, but we must re-express carefully.
Set $ u + rac{1}{u} = 2 $. Multiply by $u$: $ u^2 - 2u + 1 = 0 \Rightarrow (u - 1)^2 = 0 \Rightarrow u = 1 $, still.
So \( rac{z + w}{z - w} = 1 \Rightarrow z + w = z - w \Rightarrow
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Actually, \( u + rac{1}{u} = 2 \) implies \( u = 1 \) (since for complex \(u\), this only holds if \(|u| = 1\) and conjugate symmetry, but equality to 2 forces \(u = 1\) in \( \mathbb{C} \) only if real and positive, but let's solve generally).

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