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Exercise 2.9 · Q8

Q.If ∣z−3z∣=2\left|z-\dfrac3z\right|=2, then the least value of ∣z∣|z| is

(1) 11
(2) 22
(3) 33
(4) 55
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For fixed ∣z∣=t|z|=t, as arg⁡z\arg z varies, ∣z−3z∣\left|z-\dfrac3z\right| sweeps between ∣t−3t∣\left|t-\dfrac3t\right| and t+3tt+\dfrac3t; requiring 22 to be an attainable value forces ∣t−3t∣≤2≤t+3t\left|t-\dfrac3t\right|\le2\le t+\dfrac3t, and we solve for the smallest such tt.

Step 1. Set t=∣z∣>0t=|z|>0 and bound ∣z−3z∣\left|z-\dfrac3z\right| using the modulus inequalities.

∣∣z∣−∣3z∣∣  ≤  ∣z−3z∣  ≤  ∣z∣+∣3z∣  ⟹  ∣t−3t∣≤2≤t+3t.\left||z|-\left|\frac3z\right|\right|\;\le\;\left|z-\frac3z\right|\;\le\;|z|+\left|\frac3z\right| \;\Longrightarrow\; \left|t-\frac3t\right|\le2\le t+\frac3t.

Step 2. Check the upper bound 2≤t+3t2\le t+\dfrac3t is automatic. By AM–GM, t+3t≥23≈3.46>2t+\dfrac3t\ge2\sqrt3\approx3.46>2 for every t>0t>0, so this side never restricts tt — the binding condition is purely ∣t−3t∣≤2\left|t-\dfrac3t\right|\le2.

Step 3. Solve the boundary equations t−3t=±2t-\dfrac3t=\pm2.

t−3t=2⇒t2−2t−3=0⇒(t−3)(t+1)=0⇒t=3t-\dfrac3t=2 \Rightarrow t^2-2t-3=0 \Rightarrow (t-3)(t+1)=0 \Rightarrow t=3 (taking t>0t>0). …

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