If $|z-2+i| \leq 2$, then the difference between the greatest and least value of $|z|$ is…

If $|z-2+i| \leq 2$, then the difference between the greatest and least value of $|z|$ is $(\mathrm{i}=\sqrt{-1})$
  1. $2 \sqrt{5}+4$
  2. $2 \sqrt{5}$
  3. $4$
  4. $8$

Solution

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

Asked in: MHT CET 2023 (12 May Shift 2)

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