If $$\left|z - \frac{4}{z}\right| = 2$$, then the maximum value of $$|z|$$ is equal to

Let the complex number be written in polar form: $$z = r\,e^{i\theta}$$, where $$r = |z| \gt 0$$ and $$\theta \in [0,2\pi)$$.

Then $$\displaystyle \frac{4}{z} = \frac{4}{r}\,e^{-i\theta}$$. The given condition is

$$\left|\,z - \frac{4}{z}\,\right| = 2 \; \Longrightarrow \; \Bigl|\,r\,e^{i\theta} - \frac{4}{r}\,e^{-i\theta}\Bigr| = 2.$$

Square both sides to remove the modulus:

$$\Bigl|\,r\,e^{i\theta} - \tfrac{4}{r}\,e^{-i\theta}\Bigr|^{2} = \Bigl(r\,e^{i\theta} - \tfrac{4}{r}\,e^{-i\theta}\Bigr) \Bigl(r\,e^{-i\theta} - \tfrac{4}{r}\,e^{i\theta}\Bigr).$$

Carrying out the multiplication gives

$$r^{2} + \frac{16}{r^{2}} - 4\bigl(e^{2i\theta} + e^{-2i\theta}\bigr) = r^{2} + \frac{16}{r^{2}} - 8\cos(2\theta).$$

The squared modulus is given to be $$2^{2}=4$$, so

$$r^{2} + \frac{16}{r^{2}} - 8\cos(2\theta) = 4.$$

For a fixed $$r$$ the expression on the left is smallest when $$\cos(2\theta)=1$$ (because it is being subtracted). Hence the necessary condition for some $$\theta$$ to satisfy the equation is

$$r^{2} + \frac{16}{r^{2}} - 8 \le 4 \;\Longrightarrow\; r^{2} + \frac{16}{r^{2}} \le 12.$$

Define $$x = r^{2}\;(x \gt 0)$$. The inequality becomes

$$x + \frac{16}{x} \le 12.$$

To find the extreme values of $$x$$ that satisfy equality, solve

$$x + \frac{16}{x} = 12 \;\Longrightarrow\; x^{2} - 12x + 16 = 0.$$

Using the quadratic formula:

$$x = \frac{12 \pm \sqrt{12^{2}-4\cdot16}}{2} = \frac{12 \pm \sqrt{144-64}}{2} = \frac{12 \pm \sqrt{80}}{2} = 6 \pm 2\sqrt{5}.$$

Both roots are positive, so $$x$$ (and hence $$r^{2}$$) lies in the closed interval

$$6 - 2\sqrt{5} \;\le\; r^{2} \;\le\; 6 + 2\sqrt{5}.$$

The largest permissible modulus is therefore

$$r_{\text{max}} = \sqrt{\,6 + 2\sqrt{5}\,}.$$

Notice that $$(\sqrt{5}+1)^{2} = 5 + 1 + 2\sqrt{5} = 6 + 2\sqrt{5},$$ so

$$r_{\text{max}} = \sqrt{5} + 1.$$

Hence the maximum value of $$|z|$$ is $$\sqrt{5} + 1$$.

Option B which is: $$\sqrt{5} + 1$$