<s>[INST] Find minimum $|z|$ satisfying $|z + 1/z |= 2$. When I tried this using normal complex inequalities like $|z_{1} - z_{2}| \ge ||z_{1}| - |z_{2}||$. $\sqrt 2 - 1$ came up but the real answer seems to be $(3 - 2\sqrt 2)^{1/2}$. Some online answers on other sites support my answer as well, but I am confused which ones correct. If the latter is correct please explain how.
 [/INST] Render $\sqrt{3-2\sqrt2}=\sqrt x-\sqrt y$ for $x$ and $y$ presumed rational. Then
$3-2\sqrt2=(\sqrt x-\sqrt y)^2=(x+y)-2\sqrt{xy}$
where the quadratic surds, for rational $x$ and $y$, are equal only if the rational components and square-root components are separately equal. This leads to
$x+y=3$
$xy=2$
$x(3-x)=2, x^2-3x+2=0, x\in{1,2}$
Since $x>y$ for a positive square root $\sqrt x-\sqrt y$, we must have $x>3/2$ so we take $x=2,y=3-x=1$, and...
$\sqrt{3-2\sqrt2}=\sqrt 2-1.$
The claimed disagreement between your answers does not exist.
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