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If $|z-2+i| \leq 2$, then the difference between the greatest and least value of $|z|$ is $(\mathrm{i}=\sqrt{-1})$
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2254 Upvotes
Verified Answer
The correct answer is:
$4$
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 $=2+\sqrt{5}-(\sqrt{5}-2)=4$
$$
\begin{array}{ll}
\therefore \quad & |z-2+i| \leq 2 \\
& \Rightarrow|| z|-| 2-i|| \leq 2 \\
& \Rightarrow-2 \leq|z|-|2-i| \leq 2
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow-2 \leq|z|-\sqrt{4+1} \leq 2 \\
& \Rightarrow-2 \leq|z|-\sqrt{5} \leq 2 \\
& \Rightarrow \sqrt{5}-2 \leq|z| \leq 2+\sqrt{5}
\end{aligned}
$$
$\Rightarrow$ Largest value of $|z|$ is ' $2+\sqrt{5}$ ' and the least value is ' $\sqrt{5}-2$ '
$\therefore \quad$ Required difference $=2+\sqrt{5}-(\sqrt{5}-2)=4$
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