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Question

Mathematics Question on complex numbers

If ∣z∣≥3|z| \ge 3, then the least value of ∣z+14∣\left|z+\frac{1}{4}\right| is

A

112\frac{11}{2}

B

114\frac{11}{4}

C

33

D

14\frac{1}{4}

Answer

114\frac{11}{4}

Explanation

Solution

∣z+14∣\left|z+\frac{1}{4}\right|
=∣z−(−14)∣≥∣z∣−∣−14∣= \left|z-\left(\frac{-1}{4}\right)\right| \ge \left|z\right|-\left|\frac{-1}{4}\right|
=∣(−z)−14∣≥∣3−14∣=114= \left|\left(-z\right)-\frac{1}{4}\right|\ge \left|3-\frac{1}{4}\right|= \frac{11}{4}
Hence, ∣z+14∣≥114\left|z+\frac{1}{4}\right| \ge \frac{11}{4}