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Exercise 2.9 · Q7

Q.If ∣z−2+i∣≤2|z-2+i|\le2, then the greatest value of ∣z∣|z| is

(1) 3−2\sqrt3-2
(2) 3+2\sqrt3+2
(3) 5−2\sqrt5-2
(4) 5+2\sqrt5+2
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By the triangle inequality, for any point zz in a disk centered at cc with radius rr, the farthest such zz can be from the origin is ∣c∣+r|c|+r — reached at the point diametrically opposite the origin through the center.

Step 1. Rewrite the given condition in center–radius form.

∣z−2+i∣≤2  ⟺  ∣z−(2−i)∣≤2.|z-2+i|\le2 \;\Longleftrightarrow\; |z-(2-i)|\le2.

So zz ranges over the closed disk with center c=2−ic=2-i and radius r=2r=2.

Step 2. Apply the triangle inequality to bound ∣z∣|z|.

∣z∣=∣z−c+c∣≤∣z−c∣+∣c∣≤2+∣c∣.|z|=|z-c+c|\le|z-c|+|c|\le2+|c|. …

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