$| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$ Suppose $z=u+iv$, with $v&gt;0$.

$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ 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 $$v&gt;0,\qquad m(z)=\dfrac{-z+\sqrt{z^2-4}}{2},\qquad m(z)+z= \dfrac{z+\sqrt{z^2-4}}{2}$$

we have $$Im[m(z)+z] &gt;Im[m(z)] $$ 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 $$|m(z)+z|\ge |m(z)| $$
Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r&gt;0$

$$r^2=z^2-(z^2-r^2)=(a+ib)^2-(c+id)^2=a^2+d^2-b^2-c^2+2i(ab-cd)$$

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

$\implies a=ck,d=bk$

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

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

$\implies$ either $k&gt;1$ or $k&lt;-1$

$$|-z+\sqrt{z^2-r^2}|^2=|c+id-(a+ib)|^2=(c-a)^2+(d-b)^2=(b^2+c^2)(1-k)^2$$

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

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

So, the proposition will be nullified if $k&lt;-1$
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 $0&gt;-\Im z =\Im (m_1+m_2)=(r-\frac{1}{r})\sin \theta$ hence $r-\frac{1}{r} \le 0$ and we are done!