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If $z$ be a complex number such that

$\left|z+\dfrac{1}{z}\right|=a$

then determine the range of values that $|z|$ can take.

My Attempt: $a^2=\left|z+\dfrac{1}{z}\right|^2=\left(z+\dfrac{1}{z}\right)\left(\bar z+\dfrac{1}{\bar z}\right)$

which simplifies to

$|z|^4-(a^2+2)|z|^2+1=-(z+\bar z)^2$

Thus

$|z|^4-(a^2+2)|z|^2+1=-(z+\bar z)^2\leq 0$ (*)

$\left(|z|^2-\dfrac{a^2+2-\sqrt{a^4+4a^2}}{2}\right)\left(|z|^2-\dfrac{a^2+2+\sqrt{a^4+4a^2}}{2}\right)\leq 0$

$|z|^2\in \left[\dfrac{a^2+2-\sqrt{a^4+4a^2}}{2},\dfrac{a^2+2+\sqrt{a^4+4a^2}}{2}\right]$

$|z|\in \left[\dfrac{\sqrt{a^2+4}-a}{2},\dfrac{\sqrt{a^2+4}+a}{2}\right]$

which is a standard textbook solution.

Now the extreme values are obtained when $z+\bar z=0$ (Refer (*))

i.e. $z$ is purely imaginary.

This range is also obtained when we use the triangle inequality

$a=\left|z+\dfrac{1}{z}\right|\geq \left||z|-\dfrac{1}{|z|}\right|$.

My question is why we don't get the same answer if I use

$a=\left|z+\dfrac{1}{z}\right|\leq |z|+\dfrac{1}{|z|}$

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  • $\begingroup$ Could you use $-(z+\bar{z})^2-|z|^4\leq0$ in (*).? $\endgroup$ – Nosrati Jan 20 '17 at 16:09
  • $\begingroup$ cut-the-knot.org/arithmetic/algebra/… $\endgroup$ – lab bhattacharjee Jan 20 '17 at 16:12
  • $\begingroup$ I know. Why you didn't use a expression like $-(z+\bar{z})^2-|z|^4-1\leq0$ $\endgroup$ – Nosrati Jan 20 '17 at 16:16
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    $\begingroup$ It's probably due to $a\geq ||z|-1/|z||$ achieving equality when $z$ is purely imaginary, while $a\leq |z|+1/|z|$ achieves equality when $z$ is real. $\endgroup$ – Ennar Jan 20 '17 at 17:04
  • $\begingroup$ @Ennar.This is what my question is. Why prefer $a\geq ||z|-1/|z||$ over $a\leq |z|+1/|z|$ $\endgroup$ – Maverick Jan 21 '17 at 0:51
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A harder way of solving this question is here in my answer. If $\space\exists\space\text{z}\in\mathbb{C}$:

$$\text{z}+\frac{1}{\text{z}}=\text{z}+\frac{\overline{\text{z}}}{\text{z}\cdot\overline{\text{z}}}=\text{z}+\frac{\overline{\text{z}}}{\left|\text{z}\right|^2}=\Re\left(\text{z}\right)+\Im\left(\text{z}\right)i+\frac{\Re\left(\text{z}\right)-\Im\left(\text{z}\right)i}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag1$$

So:

  • $$\Re\left(\text{z}+\frac{1}{\text{z}}\right)=\Re\left(\text{z}\right)+\frac{\Re\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag2$$
  • $$\Im\left(\text{z}+\frac{1}{\text{z}}\right)=\Im\left(\text{z}\right)-\frac{\Im\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}\tag3$$

So, we find:

$$\left|\text{z}+\frac{1}{\text{z}}\right|=\sqrt{\Re^2\left(\text{z}+\frac{1}{\text{z}}\right)+\Im^2\left(\text{z}+\frac{1}{\text{z}}\right)}=$$ $$\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)+\frac{1+4\cdot\Re^2\left(\text{z}\right)}{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}-2}\tag4$$

In order to find the maximum and minimum:

$$ \begin{cases} \frac{\partial\space\left|\text{z}+\frac{1}{\text{z}}\right|}{\partial\space\Re\left(\text{z}\right)}=0\\ \\ \frac{\partial\space\left|\text{z}+\frac{1}{\text{z}}\right|}{\partial\space\Im\left(\text{z}\right)}=0 \end{cases}\space\space\space\space\Longleftrightarrow\space\space\space\space \begin{cases} \Re\left(\text{z}\right)\cdot\left(4\cdot\Im^2\left(\text{z}\right)+\left(\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)\right)^2-1\right)=0\\ \\ 2\cdot\Im\left(\text{z}\right)-\frac{2\cdot\Im\left(\text{z}\right)\cdot\left(1+4\cdot\Re^2\left(\text{z}\right)\right)}{\left(\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)\right)^2}=0 \end{cases}\tag5 $$

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