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Let $a> 0$. Knowing that $z$ is complex number with $|z +\frac{1}{z}|=a$ find extreme values ​​of $|z|$.

My partial solution: $$|z +\frac{1}{z}|=a \iff (z +\frac{1}{z})(\overline{z} +\frac{1}{\overline{z}})=a^2\iff |z|^2+\frac{1}{|z|^2} + (\frac{z}{\overline{z}}+\frac{\overline{z}}{z})=a^2 \iff (\frac{z}{\overline{z}}+\frac{\overline{z}}{z})=a^2 - |z|^2+\frac{1}{|z|^2}$$

It can be shown easily that $(\frac{z}{\overline{z}}+\frac{\overline{z}}{z}) \in [-2, 2].$

Using inequalities $-2\leq a^2 - |z|^2+\frac{1}{|z|^2}\leq 2$ should obtain the required result but we came to a very complicated.

Does anyone have a simple idea?

I found a solution using trigonometric form of complex numbers. This solution is further exposed, may interest someone.

$$|z +\frac{1}{z}|=a \iff |r(\cos t+i\sin t)+\frac{1}{r(\cos t+i\sin t)}|=a$$

$$\iff |r(\cos t+i\sin t)+r(\cos t-i\sin t)|=a \iff |(r+\frac{1}{r})\cos t+i(r-\frac{1}{r})\sin t|=a$$

$$\iff (r+\frac{1}{r})^2\cos^2 t+(r-\frac{1}{r})^2\sin^2 t = a^2\iff r^2+\frac{1}{r^2}+2\cos {2t}=a^2$$

$$\iff r^4-(a^2-2\cos {2t})r^2+1=0$$

Denote $r^2=u$. The roots of the equation $u^2-(a^2-2\cos {2t})u+1=0$ are $u_{1,2}=\frac{a^2-2\cos {2t}+-\sqrt{a^4-4a^2\cos {2t}-4\sin^2{2t} }}{2}= \cdots = \frac{a^2+4\sin^2 t-2+-\sqrt{(a^2-4\sin^2 t-4)(a^2-4\sin^2 t)}}{2}$

Using the notation $a^2+4\sin^2 t-2=x$ we have $u_1= \frac{x+\sqrt{x^2-4}}{2} $.

It is clear that this is maximum for $x$ max ie for $ \sin^2 t =1$ and consequently $r_{max}=\sqrt{\frac{a^2+2+\sqrt{a^4+4a^2}}{2}}= \frac{a+\sqrt{a^2+4}}{2}$ and
$r_{min}=\frac{1}{r_{max}}=\frac{\sqrt{a^2+4}-a}{2}$.

Extreme values ​​are obtained for $z_{max}=+-\frac{a+\sqrt{a^2+4}}{2}i$ and $z_{min}=+-\frac{\sqrt{a^2+4}-a}{2}i.$

(For simple writing of $r_{max}$ we used the formula

$\sqrt{A+\sqrt{B}}= \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A+C}{2}}$, where $C=\sqrt{A^2-B}$ ).

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    $\begingroup$ I recommend using \overline{z} which looks like $\overline{z}$ to denote the complex conjugate of $z$ $\endgroup$ – graydad Oct 29 '14 at 21:11
  • $\begingroup$ Longleftrightarrow=$\Longleftrightarrow$ $\endgroup$ – RE60K Oct 30 '14 at 13:34
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    $\begingroup$ @Aditya or more simply \$\iff\$ :) $\endgroup$ – Kim Jong Un Oct 30 '14 at 13:35
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You can simply use the fact that:

$|z|+\frac{1}{|z|} \geq |z+\frac1z| \geq ||z|-\frac{1}{|z|}|$

Substituting $|z+\frac1z|=a$, we get two inequalities:

$|z|+\frac{1}{|z|} \geq a$ and $a\geq ||z|-\frac{1}{|z|}|$

Taking $|z| = t$$\geq1$, we get:

$t^2 - at + 1\geq0$ and $t^2 - at - 1\leq0$

For 1st inequality let $t_1$ and $t_2$ be the roots such that $t_2 > t_1$ and similarly for 2nd inequality let $t_3$ and $t_4$ be the roots such that $t_2 > t_1$, which can be figured out using the quadratic formula (trust me, they are ugly to type).

Next, the inequalities simplify to :

$(t-t_1)(t-t_2) \geq0$ and $(t-t_3)(t-t_4) \leq0$

Getting first set as $t\in$R$-(t_1,t_2)$ and second as $t\in [t_3,t_4]$

Plotting these two on the number line along with $t>1$,

Number line We get one extremity (maximum) as $t_4$; either repeating for $t<1$ or replacing $t$ by $\frac1t$, we get the second extremity (minimum) as $\frac1t_4 $ (whose denominator can be rationalized).

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