JEE MAIN - Mathematics (2009 - No. 5)
If $$\,\left| {z - {4 \over z}} \right| = 2,$$ then the maximum value of $$\,\left| z \right|$$ is equal to :
$$\sqrt 5 + 1$$
2
$$2 + \sqrt 2 $$
$$\sqrt 3 + 1$$
Explanation
Given that $$\left| {z - {4 \over z}} \right| = 2$$
Now $$\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$$
$$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$$
$$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$$
$$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$$
$$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$$
$$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$$
$$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$$
Now $$\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}$$
$$ \Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}$$
$$ \Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0$$
$$ \Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0$$
$$\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0$$
$$ \Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)$$
$$ \Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1$$
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