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MHT CET · Maths · Complex Number

If \(|z-2+i| \leq 2\), then the difference between the greatest and least value of \(|z|\) is \((\mathrm{i}=\sqrt{-1})\)

  1. A \(2 \sqrt{5}+4\)
  2. B \(2 \sqrt{5}\)
  3. C \(4\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

Note that \(\left|z_1-z_2\right| \geq|| z_1|-| z_2||\)
\(\therefore |z-2+i| \leq 2 \)
\( \Rightarrow|| z|-| 2-i|| \leq 2 \)
\( \Rightarrow-2 \leq|z|-|2-i| \leq 2\)
\(\Rightarrow-2 \leq|z|-\sqrt{4+1} \leq 2 \)
\( \Rightarrow-2 \leq|z|-\sqrt{5} \leq 2 \)
\( \Rightarrow \sqrt{5}-2 \leq|z| \leq 2+\sqrt{5}\)
\(\Rightarrow\) Largest value of \(|z|\) is ' \(2+\sqrt{5}\) ' and the least value is ' \(\sqrt{5}-2\) '
\(\therefore\) Required difference \(=2+\sqrt{5}-(\sqrt{5}-2)=4\)