If ∣∣z−1z∣∣ = 1, then:
∣∣z−1z∣∣ = 1, then |z| = If ∣∣z−1z+1z∣∣ ≤ ∣∣z−1z∣∣ + ∣∣1z∣∣
Then |z| ≤ 1 + 1|z| ⇒ |z|2 - |z| - 1 ≤ 0
∴ |z|max = 1+√52