Krdytkyu

With $zin mathbb C$ find the maximum value for |z| such that
$$leftlvert z+frac{1}{z}rightrvert=1.$$
Source: List of problems for math-contest training.

My attempt: it is easy to see that the given condition is equivalent
$$lvert z^2+1rvert=lvert zrvert$$
and if $z=a+bi$,
begin{align*}
lvert zrvert&=lvert a^2-b^2+1+2ab irvert=sqrt{(a^2-b^2+1)^2+4 a^2b^2}\
&=sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\
&=sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\
&=sqrt{lvert zrvert^4+2(a^2-b^2)+1}
end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.

Hints and answers are welcomed.

Since $$|z| = |z^2-(-1)|geq |z|^2-|-1|$$

Write $r=|z|$ and we have $$r^2-r-1leq 0implies rin [0,r_2]$$ where $$r_{1,2} ={1pm sqrt{5}over 2} $$

So $rleq displaystyle{1+ sqrt{5}over 2}$

The reverse triangle inequality implies that $|z^2+1|geq ||z^2|-1|$ so that if $|z|^2geq 1$ we see that
$$|z|geq |z|^2-1$$
By solving this inequality using the quadratic formula, we see that $|z|leq frac{1}{2}(1+sqrt{5})$, so the maximum must be less than $frac{1}{2}(1+sqrt{5})$. We see that this maximum is attained by setting $z=frac{1}{2}(1+sqrt{5})$.

Here's a solution using the dual problem. Assume that the solution of the problem occurs at $z = re^{itheta}.$ Consider
$$
F_r(t) = |re^{it}+r^{-1}e^{-it}|^2 = r^2 + r^{-2} + e^{2it}+e^{-2it} = r^2+r^{-2}+2cos 2t.
$$ If $z= re^{itheta}$ is the maximizer, we claim that $F_r'(theta)=0$. That is, given $|z|=r$, the angle $theta$ should be chosen so that $F_r(t)$ is either maximized or minimized. To see this, assume to the contrary that $F_r'(theta)neq 0$. Then, there is small $u$ (we allow $u$ to have negative sign) such that $$F_r(theta-u)<1=F(theta)<F_r(theta+u).$$ If it is the case that $rleq 1$, then by choosing slightly large $r'$, we can make $F_{r'}(theta+u)=1$. This contradicts maximality of $r$. If $r>1$, then by choosing slightly large $r'$, we can make $F_{r'}(theta-u) = 1$. This also leads to a contradiction. Now it follows that $cos(2theta)$ should be either $-1$ or $1$. However, we can easily see that $cos(2theta)$ cannot be $1$. Thus, from $cos(2theta)=-1$, we have
$$
1=F_r(theta)=r^2 +r^{-2} -2 = (r-r^{-1})^2,
$$or
$$
r-1/r = pm 1.
$$ This gives
$$
r= frac{pm 1+ sqrt{5}}{2}.
$$ Therefore, the larger one $r=frac{ 1+ sqrt{5}}{2}$ is the solution.

Suppose $z + frac{1}{z} = e^{it}$ for some $t in mathbb{R}$. Then solving for $z$ gives
$$z = frac{e^{it} pm sqrt{e^{2it} - 4}}{2} = e^{it} cdot frac{1 pm sqrt{1 - 4 e^{-2it}}}{2}.$$
Therefore,
$$ |z| = frac{1}{2} left| 1 pm sqrt{1 - 4 e^{-2it}} right| le frac{1}{2} left( |1| + |sqrt{1 - 4 e^{-2it}}|right) = frac{1}{2} left( 1 + sqrt{1 - 4 e^{-2it}} right)le \
frac{1}{2} left( 1 + sqrt{|1| + |4 e^{-2it}|}right) = frac{1 + sqrt{5}}{2}.$$
On the other hand, if we set $t := frac{pi}{2}$ so that $e^{-2it} = -1$ then we will get equality. Therefore, the maximum value of $|z|$ is $frac{1+sqrt{5}}{2}$, which is achieved for $z = frac{1+sqrt{5}}{2} i$. (In fact, a slight refinement of the argument will show that equality holds, i.e. the maximum is achieved, if and only if $z = pm frac{1+sqrt{5}}{2} i$.)

$|z| =|z+1/z-1/z| le$

$ |z+1/z| +|1/z| =1+|1/z|;$

$r:=|z|$;

$r-1/r le 1;$

$r^2-r -1le 0;$

$(r-1/2)^2 -1/4-1le 0;$

$r-1/2 le (1/2)√5;$

$r le (1/2)(1+√5)$.

Is the maximum attained?

We can maximize $g(a,b) = a^2+b^2$ subject to $(a^2+b^2)^2 =3b^2-1$. Solve for $a^2$ we have: $a^2= sqrt{3b^2-1} -b^2implies g(a,b) = sqrt{3b^2-1}$. Thus the problem now is the find the maximum value of $b^2$. To this end, let $c = b^2 ge 0$, we have: $a^4+2a^2c+c^2 - 3c+ 1 = 0implies c^2-3c+1 le 0implies c le dfrac{3+sqrt{5}}{2}implies 3b^2 -1 = 3c-1le dfrac{7+3sqrt{5}}{2}implies |z|_{text{max}} = sqrt{g(a,b)}_{text{max}} = sqrt[4]{dfrac{7+3sqrt{5}}{2}}= dfrac{1+sqrt{5}}{2}$,and this max occurs when $a = 0$.