# $| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$

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 $$|m(z)+z|\ge |m(z)|$$

• $\sqrt {z^{2}-4}$ 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, – Kabo Murphy Nov 8 at 12:48
• $m_1(z) m_2(z)=1$ 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 – Conrad Nov 8 at 13:30
• Both tags seen unfit for your question, please read tag descriptions first. – 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 $$r>0$$

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

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

$$\implies a=ck,d=bk$$

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

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

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

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

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

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

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

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

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