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.
 A: $|z^2+1| = 2|z|$. Let $z = a+bi \implies |z^2+1| = |(a+bi)^2+1|=|a^2+2abi-b^2+1|=\sqrt{(a^2-b^2+1)^2+4a^2b^2}=2\sqrt{a^2+b^2}\implies (a^2-b^2+1)^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1-2a^2b^2+2a^2-2b^2+4a^2b^2=4a^2+4b^2\implies a^4+b^4+1+2a^2b^2-2a^2-2b^2=4b^2\implies (a^2+b^2-1)^2=4b^2\le 4(a^2+b^2)$. Let $c = a^2+b^2=|z|^2\implies (c-1)^2 \le 4c \implies c^2-2c+1 - 4c \le 0\implies c^2-6c+1\le 0\implies (c-3)^2\le 8\implies |c-3| \le 2\sqrt{2}\implies -2\sqrt{2} \le c - 3\le 2\sqrt{2}\implies c \ge 3 - 2\sqrt{2}$. Thus $|z|_{\text{min}}=\sqrt{c_{\text{min}}}=\sqrt{3-2\sqrt{2}}$. This is achieved when $a = 0, b^2-1 = \pm 2b\implies a = 0, (b\pm1)^2=2\implies a = 0, |b\pm 1| = \sqrt{2}\implies a = 0, b\pm 1=\pm \sqrt{2}\implies a = 0, b = \mp1 \pm \sqrt{2}$
A: Write $z=re^{i\theta}$. Since $|z^2+1|^2=4|z|^2$, we have a quadratic equation in $r^2$,$$r^4-r^2(4+2\cos2\theta)+1=0.$$A minimum $r^2$ takes $\cos2\theta=1$ to maximize the difference between the roots, values of $r^2$ of product $1$ (since this difference is proportional to the discriminant). So solve $r^4-6r^2+1=0$, with minimal root $r^2=3-2\sqrt{2}$.
A: 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.
A: Another way is to let $z=re^{i \theta}$, then the problem reduces to
finding the smallest $r$ that satisfies
$r^2+{1 \over r^2} + 2 (\cos^2 \theta - \sin^2 \theta) = 4$.
The graph of $f(x) = x^2 +{1 \over x^2}$ has a $\min$ at $x=1$ and $f(1) = 2$. $f$ is strictly decreasing on $(0,1]$ and strictly increasing on $[1,\infty)$.
For any $\alpha>2$ the equation $f(x) = \alpha$ has two solutions, one in $(0,1)$ and one in $(1,\infty)$.
We are interested in choosing $\theta$ such that the smallest solution of $f(r) = 4-2 (\cos^2 \theta - \sin^2 \theta)$ is a minimum. Since $f$ is strictly decreasing on $(0,1]$ we want to maximise the value of $4-2 (\cos^2 \theta - \sin^2 \theta)$ and
it is straightforward to see that the $\max$ value is $6$.
Hence we want to solve $f(r) = 6$, this can be cast as a quadratic and the smallest value of $r$ is given by
$r^2 = 3-\sqrt{8}$.
