Let $z$ be a complex number such that $$\left| z + \frac{2}{z} \right| = 2$$ Find the largest possible value of $|z|.$
I have noticed that there was a very similar version of this problem here, and illustrated the geometric approach to a problem like this. However, I'm still quite stumped and not sure how this would work in an algebraic approach. I'm assuming we can start by forming an inequality of some sort, but I'm not fully sure. Thanks in advance for the help.
The triangle inequality says:
$|z|-|\frac 2z|\le|z + \frac {2}{z}|\le |z| + |\frac 2z|$
Suppose that there is some arguement of $z$ that will make:
$|z|-\frac 2{|z|}=|z + \frac {2}{z}|= 2$ and some arguement that will make $|z| + \frac 2{|z|}=|z + \frac {2}{z}|$
In fact if $z$ is pure imaginary will be one extreme, and $z$ real will be the other.
These then become quadratics that are simple to solve.
$|z| - \frac {2}{|z|} = 2\\ |z|^2 - 2|z| - 2 = 0\\ |z| = 1 + \sqrt {3}$
To maximize $|z|$ while obeying the bound, $2/z$ should point in the opposite direction, i.e z proportional to $i$, (or $-i$). If $z=iy$, ($y$ real) then $y-\frac{2}{y}=2$. The answer is $1+3^{1/2}$.
