Suppose $z=u+iv$, with $v>0$.

$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ satisfies the equation $$m(z)+1/m(z)+z=0 $$ from which we found that $$|m(z)|=|m(z)+z|^{-1} $$

Take the branch of the square root so that the imaginary part of $m(z)$ is postive.

($m(z) $ here is the Stieljes Transformation of the Semicircle Law)

How do we show $$|m(z)|\le 1 $$ or equivalently $$|m(z)+z|\ge 1 $$

As pointed out by Conrad in the comment,$$|m(z)(z+m(z))|=1 $$If we can show $m(z)$ is the smaller of the two in norm, then we are done.

Noting that $$v>0,\qquad m(z)=\dfrac{-z+\sqrt{z^2-4}}{2},\qquad m(z)+z= \dfrac{z+\sqrt{z^2-4}}{2}$$

we have $$Im[m(z)+z] >Im[m(z)] $$ but how about the real part?

as shown in the first few lines of the accepted solution, by the constrains we have on the imaginary part of $z$ and $m(z)$. we have that the real and imaginary part of $$z$$ and $$\sqrt{z^2-4}$$ are of the of the sign. With this we can conclude that $$|m(z)+z|\ge |m(z)| $$

  • $\begingroup$ $\sqrt {z^{2}-4}$ has two values (unless $z^{2}=4$). The validity of your inequality depends on how you choose the square root. For example if $z=2i$ then inequality is not true for one of the choices of square root, $\endgroup$ – Kabo Murphy Nov 8 at 12:48
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    $\begingroup$ $m_1(z) m_2(z)=1$ so in general one of them will have an absolute value less or equal to one and the other greater or equal to one, so obviously there will be a good choice and a bad choice (except in the case where both have absolute value $1$) -maybe the question is not written here complete as you have a given choice and you need to show that works $\endgroup$ – Conrad Nov 8 at 13:30
  • $\begingroup$ Both tags seen unfit for your question, please read tag descriptions first. $\endgroup$ – Viktor Glombik Nov 8 at 15:20

Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r>0$


Equating the imaginary parts, $ab-cd=0\implies\dfrac ac=\dfrac db=k$(say)

$\implies a=ck,d=bk$

Equating the real parts, $$r^2=a^2+d^2-b^2-c^2=(b^2+c^2)(k^2-1)$$

$\implies k^2-1=\dfrac{r^2}{b^2+c^2}>0$

$\implies$ either $k>1$ or $k<-1$


$$=\dfrac{r^2(1-k)^2}{(k^2-1)}=\dfrac{r^2(k-1)}{k+1}$$ which will be $\le r^2$

$\iff\dfrac{k-1}{k+1}\le1\iff0\ge\dfrac{k-1}{k+1}-1=-\dfrac2{k+1}$ $\iff 0\le\dfrac2{k+1}\iff k+1>0\iff k>-1$

So, the proposition will be nullified if $k<-1$

  • $\begingroup$ by assumption $b>0$, and the square root is defined to make $Im(m(z)>0)$ which implies $d>0$, so $d/b>0$. $\endgroup$ – a point in Standard Students Nov 9 at 20:35
  • $\begingroup$ but $a/c=d/b>0 $ solves my problem $\endgroup$ – a point in Standard Students Nov 9 at 20:36

Let's write $m_1,m_2$ the 2 roots of the equation $m^2+zm+1=0$ where $\Im m_1 \ge 0$ so $m_1=m(z)$ in the post notation.

We know that $m_1=re^{i\theta}, 0 \le \theta \le \pi$, so $\sin \theta \ge 0$ while then $m_2=\frac{1}{r}e^{-i\theta}$, so $0>-\Im z =\Im (m_1+m_2)=(r-\frac{1}{r})\sin \theta$ hence $r-\frac{1}{r} \le 0$ and we are done!

  • $\begingroup$ $m_2$ will be the conjuagte of $m_1$, if this is the case shouldn't $m_2=re^{-i\theta}$ $\endgroup$ – a point in Standard Students Nov 9 at 19:28
  • $\begingroup$ @a point no that is not true since $z$ is not real so the roots are not conjugate $\endgroup$ – Conrad Nov 9 at 21:14
  • $\begingroup$ @ Conrad I see. The part I don't understand is why $m_2=\dfrac{1}{r}e^{-i\theta}$, $m_2=m_1^{-1}$ $\endgroup$ – a point in Standard Students Nov 9 at 21:17
  • $\begingroup$ $m_1m_2=1$ by the usual viete relations (roots monic quadratic with free term $1$ $\endgroup$ – Conrad Nov 9 at 23:16

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