# Inequality in the complex plane

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.

• Define $\sqrt {z^{2}+a}$ Sep 7, 2021 at 23:48
• @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.
– Goga
Sep 7, 2021 at 23:52
• Your first line is pretty close. It already looks like $C / |z|$. so can you force it into that form? Sep 7, 2021 at 23:55
• @CalvinLin for $\Re(z)$ positive I think that it is true?. Did you find my proof incorrect?
– Goga
Sep 7, 2021 at 23:58
• Personally speaking, your proof isn't the "right" to think about this, so I didn't check the details. Sep 8, 2021 at 0:01

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

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

Corollary:

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

• Thank you sir, but what about the case z is real negative?. This inequality will be false. Have we precise the Damaine of z ?
– Goga
Sep 8, 2021 at 0:29
• 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. Sep 8, 2021 at 0:31
• 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?
– Goga
Sep 8, 2021 at 0:41