If $\left| z-\dfrac{4}{z} \right|=2$, then the maximum value of $\left| z \right|$ is equal to
(a) $1+\sqrt{3}$
(b) $1+\sqrt{5}$
(c) $1-\sqrt{5}$
(d) $\sqrt{5}-1$

Hint: Use the inequalities of two complex numbers. It is given as $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right|$ and $\left| {{z}_{1}}+{{z}_{2}} \right|\le \left| \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right| \right|$. Use any one of the inequality to get inequality in $\left| z \right|$. Now, get the possible range of values of $\left| z \right|$. And hence get maximum value of $\left| z \right|$.

Complete step-by-step answer:

We know the relation between the two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$can be given as
$\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right|$ …………………… (i)
Let the complex numbers in the above equation be z and $\dfrac{4}{z}$, so that we can use the given relationship in the problem. Hence, we get
$\left| z-\dfrac{4}{z} \right|\ge \left| \left| z \right|-\dfrac{4}{\left| z \right|} \right|$…………… (ii)
Now, it is given that value of $\left| z-\dfrac{4}{z} \right|$ is 2 from the problem;
$\left| z-\dfrac{4}{z} \right|=2$ …………… (iii)
Now, we can put value of $\left| z-\dfrac{4}{z} \right|$from equation (iii) in the equation (ii) and hence, we get
$2\ge \left| \left| z \right|-\dfrac{4}{\left| z \right|} \right|$
Let us suppose $\left| z \right|='r'$ , hence, we get:
$\left| r-\dfrac{4}{r} \right|\le 2$…………………….. (iv)
Now, we know the property of inequalities in modular function, can be given as
If $\left| x \right|\le a$, then
$-a\le x\le a$ ………….. (v)
Hence, we can simplify equation (iv) as,
$-2\le r-\dfrac{4}{r}\le 2$
Now, we can solve left and right inequalities to get the range of value of ‘r’.
As $\left| z \right|=r$is a positive value; then we can multiply the term ‘r’ to the other side of the inequalities.
Hence, the left inequality gives
$\begin{align}
  & -2\le \dfrac{{{r}^{2}}-4}{r} \\
 & -2r\le {{r}^{2}}-4 \\
 & \Rightarrow {{r}^{2}}+2r-4\ge 0...........(vi) \\
\end{align}$
Now, we can find the corresponding roots of ${{r}^{2}}+2r-4=0$ with the help of quadratic formula given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for $a{{x}^{2}}+bx+c=0$
Hence, roots of ${{r}^{2}}+2r-4=0$ is given as
$r=\dfrac{-2\pm \sqrt{4+16}}{2}=\dfrac{-2\pm \sqrt{20}}{2}$
$r=\dfrac{-2\pm 2\sqrt{5}}{2}=-1\pm \sqrt{5}$ …………………… (vii)
Hence, from equation (vi) and (vii), we get
$\left( r-\left( -1+\sqrt{5} \right) \right)\left( r-\left( -1-\sqrt{5} \right) \right)\ge 0$
Hence, $r\le -1-\sqrt{5}$and $r\ge -1+\sqrt{5}$
$\Rightarrow r\in \left( -\infty ,-1-\sqrt{5} \right)\bigcup \left( \sqrt{5}-1,\infty \right)$ …………… (viii)
Again considering the right inequality, we get
$\begin{align}
  & r-\dfrac{4}{r}\le 2 \\
 & \dfrac{{{r}^{2}}-4}{r}\le 2 \\
 & {{r}^{2}}-4\le 2r \\
 & \Rightarrow {{r}^{2}}-2r-4\le 0.............(ix) \\
\end{align}$
Now, roots of the equation ${{r}^{2}}-2r-4=0$ can be given as
$\begin{align}
  & r=\dfrac{2\pm \sqrt{4+16}}{2} \\
 & r=\dfrac{2\pm \sqrt{20}}{2}=\dfrac{2\pm 2\sqrt{5}}{2} \\
 & r=1\pm \sqrt{5} \\
\end{align}$
Hence, equation (ix) can be written as
$\left( r-\left( 1+\sqrt{5} \right) \right)\left( r-\left( 1-\sqrt{5} \right) \right)\le 0$
Hence,
$1-\sqrt{5}\le r\le 1+\sqrt{5}$ …………… (x)
Now, taking intersection of inequalities of equation (viii) and (x), we get
$\sqrt{5}-1\le r\le \sqrt{5}+1$
Hence, the greatest value of r i.e. $\left| z \right|$ is $\sqrt{5}+1$
So, option (b) is the correct answer.

Note: One may try to put z = x + iy in the given problem but that would be a very complex approach for the problem and will take a lot of time.
Inequalities $\left| {{z}_{1}}-{{z}_{2}} \right|\ge \left| \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right|$ and $\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\ge \left| \left| {{z}_{1}}+{{z}_{2}} \right| \right|$ will always be a key point for all these types of the given problems. One can solve the inequality $\left| r-\dfrac{4}{r} \right|\ge 2$ by squaring both the sides and simplify it further.