Modulus represents the distance from Origin.

We know that from triangle inequality

$\left| z \right| -\left| \frac { 2 }{ z } \right| \le \left| z+\frac { 2 }{ z } \right| =2\le \left| z \right| +\left| \frac { 2 }{ z } \right|$

Now the upper limit offers no useful result as it is always true by AM-GM

The lower limit however gives the quadratic $\left| z \right| -\left| \frac { 2 }{ z } \right| -2=0\\ \left| z \right| ^{ 2 }-2\left| z \right| -2=0\\$

whose positive root is $1+\sqrt { 3 }$

(we neglect negative root as modulus cant be negative, and so no extrema can be found there)

$let\quad z=r{ e }^{ i\theta }\quad \\ \\ \Rightarrow z+\cfrac { 2 }{ z } =r{ e }^{ i\theta }+\cfrac { 2 }{ r } { e }^{ -i\theta }=\left( r+\cfrac { 2 }{ r } \right) cos\theta +i\left( r-\cfrac { 2 }{ r } \right) sin\theta \\ \\ \Rightarrow { 4=\left| z+\cfrac { 2 }{ z } \right| }^{ 2 }={ r }^{ 2 }+\cfrac { 4 }{ { r }^{ 2 } } +4cos2\theta \\ \\ \because \quad { r }^{ 2 }+\cfrac { 4 }{ { r }^{ 2 } } \ge 4(A.M-G.M)\& \quad cos2\theta \in \left[ -1,1 \right] \\ \\ \therefore { r }^{ 2 }+\cfrac { 4 }{ { r }^{ 2 } } \quad should\quad be\quad within\quad [4,8]\quad to\quad satisfy\quad above\quad { eq }^{ n }.\\ \\ Graph\quad of\quad y=x^{ 2 }+\cfrac { 4 }{ { x }^{ 2 } } for\quad x\ge 0\quad will\quad be\quad decreasing\quad in\quad \\ \\ \left( 0,\sqrt { 2 } \right) \quad and\quad increasing\quad in\left( \sqrt { 2 } ,\infty \right) .\\ \\ It\quad is\quad clear\quad from\quad above\quad statement\quad that\quad { r }_{ max }\quad will\quad be\\ \\ greater\quad than\quad \sqrt { 2 } satisfying,\\ \\ { r }^{ 2 }+\cfrac { 4 }{ { r }^{ 2 } } =8\Rightarrow { r }^{ 2 }=4\pm 2\sqrt { 3 } \\ \\ \Rightarrow \quad { r }=\pm \left( \sqrt { 3 } \pm 1 \right) \Rightarrow { r }_{ max }=\sqrt { 3 } +1=2.732\\ \\$