# Complex inequality question

I am trying to understand why the following holds: \begin{align*} \Re((1-\imath)(A+B)) \geq \Re((1-\imath)A) - \sqrt{2}|B|, \end{align*} where, \begin{align*} A:= \sum_{x=1}^{[\sqrt{k}]} e\left(\frac{x^2}{4k}\right) \end{align*} and \begin{align*} B:= \sum_{x=[\sqrt{k}]+1}^{k-1} e\left(\frac{x^2}{4k}\right), \end{align*} where $k$ is an odd positive integer, $\displaystyle e(\alpha):=e^{2\pi \imath \alpha}$ and $[k]$ is the greatest integer $\leq k$. This inequality should amount to showing \begin{align*} \Re((1-\imath)B) \geq -\sqrt{2}|B|. \end{align*} I sort of think I should use a complex modulus inequality, but I can't see it.

I am able to show that $|B| \leq \sqrt{k}$ and $\Re((1-\imath)A) > \frac{\sqrt{k}}2$. So if the first inequality holds, it follows that \begin{align*} \Re((1-\imath)(A+B)) \geq \sqrt{k}\left(\frac 12 - \sqrt{2}\right) > -\sqrt{k}. \end{align*}

This question comes from T. Estermann's paper $\textit{On the sign of the Gaussian Sum}$ given in the Journal of the London Mathematical Society, Volume 20, 1945, pp. 66-67.

## 1 Answer

If I were to write $B = x + \imath y$ then $\Re((1-\imath)(x+\imath y)) = x+y$. And we have $x+y \geq \sqrt{x^2+y^2} = |B|$. So as $\Re((1-\imath)B) \geq |B| \geq -\sqrt{2}|B|$ we have our inequality.