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$$H = \{z : z \in \mathbb{C}^*, \ \ \left\lvert z + \frac1z \right\rvert = a \}$$ Find the minimum and maximum value of $|z|$ such that $z \in H$

$$a^2 = \left\lvert z + \frac1z \right\rvert^2 = \left(z + \frac1z\right)\left(\overline z + \frac1{\overline z} \right) = |z|^2 + {z^2 + \overline z^2 \over |z|^2} + {1 \over |z|^2} = {|z|^4 - |z|^2(2) + (z + \overline z)^2 + 1 \over |z|^2 }$$

Hence, $$|z|^4 - |z|^2 \cdot(a^2 + 2) + 1 = -(z + \overline z)^2 \le 0$$

Thus,

$$|z|^2 \in \left[{a^2 + 2 - \sqrt{a^4 + 4a^2}\over 2}, {a^2 + 2 + \sqrt{a^4 + 4a^2}\over 2}\right] \tag{3}$$

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

  • How did we get from (3) to (4) ?

2 Answers2

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It's straightforward: square the limits for $|z|$ (as in (4)) to arrive at the limits for $|z|^2$ (as in (3)).

All steps for the left limit:

You need to establish $\sqrt {a^2 + 2 - \sqrt{a^4 + 4a^2}\over 2} = {-a + \sqrt{a^2 + 4}\over 2}$. Squaring this translates into

$${a^2 + 2 - \sqrt{a^4 + 4a^2}\over 2} = ({-a + \sqrt{a^2 + 4}\over 2})^2 = {a^2 -2 a \sqrt{a^2 + 4} + a^2 + 4\over 4} = {a^2 + 2 - \sqrt{a^4 + 4a^2}\over 2}$$

Done.

Likewise for the right limit.

Andreas
  • 15,175
1

The simplification follows from the more general radical denesting technique:

$$ \sqrt{u \pm \sqrt{v}} = \sqrt{\frac{u+w}{2}} \pm \sqrt{\frac{u-w}{2}} \quad\;\;\text{where}\;\; w^2 = u^2-v $$

In the case here, $u=a^2+2, v=a^4+4a^2, w^2=(a^2+2)^2-(a^4+4a)=4\,$, then:

$$ \sqrt{{a^2 + 2 \pm \sqrt{a^4 + 4a^2}\over 2}} = \frac{1}{\sqrt{2}}\left(\sqrt{\frac{a^2+4}{2}} \pm \frac{\sqrt{a^2}}{2}\right)=\frac{\sqrt{a^2+4} \pm a}{2} $$

dxiv
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