I'm stuck in the proof of the following inequality for an arbitrary $z\in \mathbb{C}$:

$$ \left\vert \sqrt{z^{2}+a}-z\right\vert \leq \frac{C}{\left\vert z\right\vert },\text{ for large }\left\vert z\right\vert , $$

where $a\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R}$ and assume that $\Re(z)$ is bounded.

If we multiply $\sqrt{z^{2}+a}-z$ by its conjugate we find $$ \sqrt{z^{2}+a}-z=\frac{a}{\sqrt{z^{2}+a}+z}. $$

If $\Re(z)\geq 0,$ we know that $\sqrt{z^{2}+a}=\sqrt{z^{2}}\sqrt{% z^{2}+a}=z\sqrt{1+\frac{a}{z^{2}}},$ therefore $$ \left\vert \frac{a}{\sqrt{z^{2}+a}+z}\right\vert \leq \left\vert \frac{1}{z}% \right\vert \frac{1}{\left\vert \sqrt{1-\left\vert \frac{a}{z^{2}}% \right\vert }+1\right\vert }\leq \frac{C}{\left\vert z\right\vert }\text{ for large }\left\vert z\right\vert . $$

Now, if $\Re(z)\leq 0$ we have $\sqrt{z^{2}+a}=\sqrt{z^{2}}\sqrt{% z^{2}+a}=-z\sqrt{1+\frac{a}{z^{2}}}$ and so $$ \sqrt{z^{2}+a}-z=-z\sqrt{1+\frac{a}{z^{2}}}-z. $$

The above approach doesn't seem to work. Any Ideas?.

Thank you.

  • $\begingroup$ Define $\sqrt {z^{2}+a}$ $\endgroup$ Sep 7, 2021 at 23:48
  • $\begingroup$ @KaviRamaMurthy if $a$ is positive, this function is defined on the branch $\mathbb{C} - [-ai,ai] $, if $a$ is negative then it is defined on the branch $\mathbb{C} - [-a,ai]$. I have added the assumption that $\Re(z)$ is bounded. $\endgroup$
    – Goga
    Sep 7, 2021 at 23:52
  • $\begingroup$ Your first line is pretty close. It already looks like $ C / |z| $. so can you force it into that form? $\endgroup$
    – Calvin Lin
    Sep 7, 2021 at 23:55
  • $\begingroup$ @CalvinLin for $\Re(z)$ positive I think that it is true?. Did you find my proof incorrect? $\endgroup$
    – Goga
    Sep 7, 2021 at 23:58
  • $\begingroup$ Personally speaking, your proof isn't the "right" to think about this, so I didn't check the details. $\endgroup$
    – Calvin Lin
    Sep 8, 2021 at 0:01

1 Answer 1


Hint: Find a large enough $ n$, so that for $ |z| > n$, (say)

$$ |z| \leq \left| \sqrt{ z^2 + a } + z \right|.$$


$$ | \sqrt{ z^2 + a } - z | \leq \frac{a} { | \sqrt{z^2 + a } + z| } \leq \frac{a}{|z|} $$

  • $\begingroup$ Thank you sir, but what about the case z is real negative?. This inequality will be false. Have we precise the Damaine of z ? $\endgroup$
    – Goga
    Sep 8, 2021 at 0:29
  • $\begingroup$ Note that $ |z| > n$, where $n$ is large enough and has to be determined. This is what makes the inequality true (IE The cases where it's false, have a small $|z|$ value.) Roughly (handwaving) speaking, if $|z|$ is very large compared to $a$, then $ \sqrt{ z^2 + a } \approx z$ and so $ | \sqrt{ z^ 2 + a } + z | \approx 2z$. We are then allowing for (more than) enough of a buffer. $\endgroup$
    – Calvin Lin
    Sep 8, 2021 at 0:31
  • $\begingroup$ Thank you sir for your kindness. Does this still true if we take $z=x<0$. I think that the inequality will be false. Must we specify the domain? $\endgroup$
    – Goga
    Sep 8, 2021 at 0:41

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