Question

If we have an expression as $\left| {z - \dfrac{4}{z}} \right| = 2$, then the maximum value of $\left| z \right|$ is
1) $\sqrt 3 + 1$
2) $\sqrt 5 + 1$
3) $2$
4) $2 + \sqrt 2 $

Answer 1

Hint: To solve the problem we will first have to use the property of complex numbers that is $\left| {\left| \alpha \right| - \left| \beta \right|} \right| \leqslant \left| {\alpha - \beta } \right|$
Using this property we will obtain an inequality with $\left| z \right|$. Then assuming $\left| z \right|$ as a positive number we can solve the inequality and will obtain two quadratic inequalities. On solving those inequations we will obtain our required result.

Complete step-by-step solution:
We are given, $\left| {z - \dfrac{4}{z}} \right| = 2$.
Now, applying the property of complex numbers, that is, $\left| {\left| \alpha \right| - \left| \beta \right|} \right| \leqslant \left| {\alpha - \beta } \right|$, we get,
$\left| {\left| z \right| - \left| {\dfrac{4}{z}} \right|} \right| \leqslant \left| {z - \dfrac{4}{z}} \right|$
$ \Rightarrow \left| {\left| z \right| - \left| {\dfrac{4}{z}} \right|} \right| \leqslant 2$
[Given, $\left| {z - \dfrac{4}{z}} \right| = 2$]
$ \Rightarrow \left| {\left| z \right| - \dfrac{4}{{\left| z \right|}}} \right| \leqslant 2$
Now, let us assume, $\left| z \right| = r > 0$, since we are taking the mod value, it will be positive.
Therefore, $ \Rightarrow \left| {r - \dfrac{4}{r}} \right| \leqslant 2$
Now, splitting the mod function over the inequality, we get,
$ \Rightarrow - 2 \leqslant r - \dfrac{4}{r} \leqslant 2$
Now, from the left inequality it gives,
$ - 2 \leqslant r - \dfrac{4}{r}$
Multiplying $r$ on both sides of the inequality, we get,
$ \Rightarrow - 2r \leqslant {r^2} - 4$
Adding $2r$ on both sides of the inequality, we get,
$ \Rightarrow 0 \leqslant {r^2} + 2r - 4$
Now, changing the side, we get,
$ \Rightarrow {r^2} + 2r - 4 \geqslant 0$
The corresponding roots are,
$ \Rightarrow r \geqslant \dfrac{{ - 2 \pm \sqrt {4 - 4\left( { - 4} \right)} }}{2}$
$ \Rightarrow r \geqslant \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{2}$
$ \Rightarrow r \geqslant \dfrac{{ - 2 \pm \sqrt {20} }}{2}$
$ \Rightarrow r \geqslant \dfrac{{ - 2 \pm 2\sqrt 5 }}{2}$
Taking, $2$ common from numerator and diving by the denominator, we get,
$ \Rightarrow r \geqslant - 1 \pm \sqrt 5 $
Therefore, $r \geqslant - 1 - \sqrt 5 $ or $r \geqslant \sqrt 5 - 1$
Since, $r > 0$.
So, $r \geqslant \sqrt 5 - 1 - - - \left( 1 \right)$.
Now, from the right inequality it gives,
$r - \dfrac{4}{r} \leqslant 2$
Multiplying $r$ on both sides of the inequality, we get,
$ \Rightarrow {r^2} - 4 \leqslant 2r$
Subtracting $2r$ on both sides of the inequality, we get,
$ \Rightarrow {r^2} - 2r - 4 \leqslant 0$
The corresponding roots are,
$ \Rightarrow r \leqslant \dfrac{{2 \pm \sqrt {4 - 4\left( { - 4} \right)} }}{2}$
$ \Rightarrow r \leqslant \dfrac{{2 \pm \sqrt {4 + 16} }}{2}$
$ \Rightarrow r \leqslant \dfrac{{2 \pm \sqrt {20} }}{2}$
$ \Rightarrow r \leqslant \dfrac{{2 \pm 2\sqrt 5 }}{2}$
Taking, $2$ common from numerator and diving by the denominator, we get,
$ \Rightarrow r \leqslant 1 \pm \sqrt 5 $
So, $r \leqslant 1 - \sqrt 5 $ or $r \leqslant 1 + \sqrt 5 $
Since, $r > 0$.
Therefore, $r \leqslant 1 + \sqrt 5 - - - \left( 2 \right)$.
Therefore, we can say from $\left( 1 \right)$ and $\left( 2 \right)$,
$\sqrt 5 - 1 \leqslant r \leqslant \sqrt 5 + 1$
$ \Rightarrow \sqrt 5 - 1 \leqslant \left| z \right| \leqslant \sqrt 5 + 1$
Therefore, the greatest value of $\left| z \right|$ is $\sqrt 5 + 1$, correct option is 2.

Note: Complex numbers are the most wide field of sets of numbers. It comprises all kinds of number sets like natural numbers, integers, real numbers, rational and irrational numbers. The complex numbers comprises of two parts, imaginary and real parts and are written in the form of $a + ib$, where $a$ is the real part and $ib$ is the imaginary part and $i$ (called iota) has the value $\sqrt { - 1} $.