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Complex Numbers Part 2

Section: Mathematics

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Question : 89 of 106

Marks: +1, -0

If

z \neq 0

be a complex number such that

\left|z-\;\frac{1}{z}\right|=2

, then the maximum value of

|z|

[29-Jul-2022-Shift-2]

$\sqrt{2}$
1
$\sqrt{2}-1$
$\sqrt{2}+1$

Solution:

We know,

\left| \left| z_1 \right| - \left| z_2 \right| \right| \leq \left| z_1+z_2 \right| \leq \left| z_1 \right| + \left| z_2 \right|

\therefore \left| \left| z \right| - \frac{1}{|z|} \right| \leq \left| z - \frac{1}{z} \right|

\Rightarrow \left| \left| z \right| - \frac{1}{|z|} \right| \leq 2 \left[ \text{ Given } \left| z - \frac{1}{z} \right| =2 \right]

\Rightarrow \left| \frac{|z|^2-1}{|z|} \right| \leq 2

\Rightarrow -2 \le \frac{|z|^2-1}{|z|} \le 2

\therefore \frac{|z|^2-1}{|z|} \le 2

\Rightarrow |z|^2-1 \le 2|z|

\Rightarrow |z|^2-2|z|-1 \le 0

\Rightarrow |z|^2-2|z|+1-2 \le 0

\Rightarrow (|z|-1)^2-2 \le 0

\Rightarrow -\sqrt{2} \le |z|-1 \le \sqrt{2}

\Rightarrow 1-\sqrt{2} \le |z| \le 1+\sqrt{2} \qquad (1)

or

-2 \le \frac{|z|^2-1}{|z|}

\Rightarrow |z|^2-1 \le -2|z|

\Rightarrow |z|^2+2|z|-1 \le 0

\Rightarrow |z|^2+2|z|+1-2 \le 0

\Rightarrow (|z|+1)^2-2 \le 0

\Rightarrow -\sqrt{2} \le |z|+1 \le +\sqrt{2}

\Rightarrow -\sqrt{2}-1 \le |z| \le \sqrt{2}-1 \qquad \text{(2)}

From (1) and (2) we get,

Maximum value of

|z| = \sqrt{2}+1

and minimum value of

|z| = -\sqrt{2}-1

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