I am doing some project. I have to calculate the estimates of an operator. But I was stuck on a part. I need to show the following form of inequality to derive a conclusion what I want to show.

$|\sqrt{z_{1}} - \sqrt{z_{2}}| < C|\sqrt{z_{1} - z_{2}}|$ for a constant $C \in \mathbb{R}$ and $z_{1}, z_{2} \in \mathbb{C}$

Does anyone give me some comments related to this inequality? $C$ might depend on the values of $z_{1}$ and $z_{2}$.

Here, $z_{1}$ and $z_{2}$ are values of complex functions and so I am not sure how I can deal with this property. I know, for the real-valued, it is trivial because $f(x) = \sqrt{x}$ is a concave function. But, how can I define a concavity of a complex function?

I scanned Functions of One Complex Variables 1 by J. B. Conway, but I couldn't find any related theorems to this case.


  • 1
    $\begingroup$ What branch cut for the square root? $\endgroup$ May 11, 2013 at 20:01

2 Answers 2


As you probably know, there is no continuous branch of $\sqrt{z}$ on the complex plane. No matter which branch we take, it will involve a cut somewhere in the plane, for example along the negative real axis. If $z_1$ and $z_2$ lie near each other on different sides of the cut, then $|\sqrt{z_1-z_2}|=\sqrt{|z_1-z_2|}$ will be small while $|\sqrt{z_1}-\sqrt{z_2}|$ will be approximately $2\sqrt{|z_1|}$. So the stated inequality cannot work for a fixed branch independent of $z_1,z_2$.

However, we can approach the problem differently: pick a branch of $\sqrt{z}$ at $z_1$ and extend it to $z_2$ in a reasonable way. To this end, let $\zeta=z_2/z_1$ and define $\sqrt{\zeta}$ to be the branch with $\sqrt{1}=1$ in $\mathbb C\setminus (-\infty,0]$. (Annoyingly, $\zeta$ could be negative; let's take $\sqrt{-1}=i$ in this case.) Then it's reasonable to set $\sqrt{z_2}=\sqrt{z_1}\sqrt{\zeta}$

Now, the question reduces to showing that $$ \left |1-\sqrt{\zeta}\right|\le C\sqrt{|1-\zeta|} \tag1$$ Since you allow for $C$ here, I will not sweat too much over sharpness, although $C=1$ should work.

  • When $|\zeta|\le 1/2$, inequality (1) holds with $C=\sqrt{2}+1$ because we have at most $1+\sqrt{1/2}$ on the left and at least $\sqrt{1/2}$ on the right.
  • When $|\zeta|\ge 1/2$, the function $$f(\zeta)=(1-\sqrt{\zeta})^2=1+\zeta-2\sqrt{\zeta}$$ satisfies $|f'(\zeta)|\le 1+1/\sqrt{|\zeta|}\le 1+\sqrt{2}$. Every point of the domain $\{|\zeta|\ge 1/2\}$ can be connected to $1$ by a curve of length at most $\pi|\zeta-1|$ (rough estimate). Integrating $f'$ along this curve, we obtain
    $$|f(\zeta) |\le (1+\sqrt{2})\pi |\zeta-1|$$ and therefore (1) holds with $C=\sqrt{(1+\sqrt{2})\pi}$.
  • $\begingroup$ I am really sorry that I checked your answer. I highly appreciate you for giving useful comments. Thanks. $\endgroup$
    – Philo
    Oct 18, 2014 at 22:54

I purpose a strategy of finding where the inequality has equality and then just testing a point to see if the inequality holds in a particular connected region.

For example,

If $\sqrt{z}-\sqrt{w}=\sqrt{z-w}$ then $z-2\sqrt{zw}+w=z-w$ and so $w^2=wz$ thus $w=z$ or $w=0$. Given a region $\Omega\subseteq\mathbb{C}^2$ such that $w\not=z$ and $w\not =0$ for all $w,z\in \Omega$, if $\Omega$ is connected then the inequality will hold entirely in $\Omega$ or hold nowhere in $\Omega$.


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