Find minimum $|z|$ satisfying $|z + 1/z |= 2$.

Question

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.

Answer 1

$|z^2+1| = 2|z|$. 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 $a = 0, b^2-1 = \pm 2b\implies a = 0, (b\pm1)^2=2\implies a = 0, |b\pm 1| = \sqrt{2}\implies a = 0, b\pm 1=\pm \sqrt{2}\implies a = 0, b = \mp1 \pm \sqrt{2}$

Answer 2

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 $r^2=3-2\sqrt{2}$.

Answer 3

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.

Answer 4

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 $r^2 = 3-\sqrt{8}$.

Answer 5

The same conclusion presented in other answers can be found naturally enough and with minimal calculations once one makes a simple conjecture with e.g. a graphing tool.Although it might be a bit convoluted, and tedious to formalize, I think the following solution can still be of interest as I didn't find it anywhere. In the following paragraphs, complex numbers will be identified with their image in the complex plane (it's not strictly the same thing but allows for lighter writing).

It can actually be shown that the set of complex numbers $z$ satisfying $\left|z+\frac{1}{z}\right|=2$ is the union of two circles in the complex plane. Now, that information is enough to determine their centers and radii, from which finding the closest point(s) to the origin is straightforward.

1.Start by assuming what we want to show (*), namely that $S=C_1\cup C_2$ where $S$ is our set and $C_1$, $C_2$ are two circles in the complex plane (actually sets of complex numbers). Let $z_1,r_1,z_2,r_2$ be the center and radius of $C_1, C_2$ respectively. Notice that if $z \in S$ then:

• its complex conjugate $\bar z$ also belongs to S
• $-z \in S$
• $-\bar z \in S$

Graphically speaking, this means the image of S in the complex plane has a four-fold reflection symmetry. This drastically reduces your choice for the centers of both circles: if the center of one of them were to lie in the first quadrant , then, by the symmetry of the figure four circles with different centers would be included in $S$ , which is the union of two circles.
It can be shown formally that this is impossible (see Sets of circles ). Therefore, this center $z_1$ must belong to either the real axis, or the imaginary axis (or both if it's $0$).

Assume for the sake of contradiction that $z_1 \in \mathbb{R}$. By way of symmetry, we also have $z_2 \in \mathbb{R}$. Each circle must have two points of intersection with the real axis. Hence, if they are not the same circle (i.e. $z_1=z_2$), there are four different points in $S$ that also belong to the real line. Let $r \in S\cap \mathbb{R}$.
We have $\left|r+\frac{1}{r}\right|=2 \Longleftrightarrow \left(r+\frac{1}{r}\right)²=4 \Longleftrightarrow \left( \left(r+\frac{1}{r}\right)-2\right)\left( \left(r+\frac{1}{r}\right)+2\right)=0 $ $\Longleftrightarrow r^2-2r+1=0 \text{ or } r^2+2r+1=0$ $\Longleftrightarrow (r-1)^2=0 \text{ or } (r+1)^2=0$ $\Longleftrightarrow r \in \{-1,1\}$.

As we only found two intersection points, this means the two circles are in fact one. This forces $z_1=z_2=0$, then $r_1=r_2=1$ meaning $C_1$ and $C_2$ both are the unit circle. But $\left|i+\frac{1}{i}\right|=0\neq 2$ so $i\notin S$: a contradiction. Therefore, $z_1 \in i\mathbb{R}$ and similarly $z_2 \in i\mathbb{R}$.
Let's apply our previous reasoning to the imaginary line. Let $z \in S\cap i\mathbb{R}$. $z$ can be written as $iy$ where $y \in \mathbb{R}$. Then $\left|z+\frac{1}{z}\right|=2 \Longleftrightarrow \left|iy+\frac{1}{iy}\right|=2 \Longleftrightarrow \left|i(y-\frac{1}{y})\right|=2 \Longleftrightarrow \left|(y-\frac{1}{y})\right|=2 \Longleftrightarrow \left(y-\frac{1}{y}\right)²-2=0 \Longleftrightarrow \left[y-\frac{1}{y}-\sqrt{2}\right]\left[y-\frac{1}{y}+\sqrt{2}\right]=0 $
Solving the quadratic in each factor gives $y\in \{-1-\sqrt{2}, 1-\sqrt{2},\sqrt{2}-1,1+\sqrt{2} \}$. These values are sorted on the real line and are the imaginary parts of the intersection points we are looking for. Then, we must have $\{z_1,z_2\}=\{ -i,i\}$ and $r_1=r_2=\sqrt{2}$

2. We have found the centers and radii of $C_1$ and $C_2$ under the assumption that $S=C_1\cup C_2$. This assumption is now easy to check by a little calculation left as an exercise. As the two circles are symmetric, it is sufficient to determine the $z\in C_1$ with the least magnitude. Assume WLOG that $z_1$ is the center of $C_1$ so that $C_1= \{z_1+\sqrt{2}e^{i\theta},\theta \in [0;2\pi]\}$.

Let $\theta \in [0;2\pi]$. The triangle inequality yields $\sqrt{2}-1 \leq |i+\sqrt{2}e^{i\theta}|$ with equality iff $\theta=-\frac{\pi}{2}$.

(*)This might seem mathematically unsound but is a frequent model of reasoning for which I can't find a translation:the idea is to find necessary conditions and gather more information- here the center and radii of the circles).