# Square roots

## Definitions

Symbol: Sqrt $\sqrt{z}$ Principal square root

## Illustrations

Image: X-ray of $\sqrt{z}$ on $z \in \left[-3, 3\right] + \left[-3, 3\right] i$ ## Elementary functions

$\sqrt{z} = {e}^{1 / 2 \log(z)}$
$\sqrt{z} = {z}^{1 / 2}$

## Specific values

Table of $\sqrt{n}$ to 50 digits for $0 \le n \le 50$
$x \in \left[0.707106781186547524400844362105 \pm 1.51 \cdot 10^{-31}\right]\; \text{ where } x = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
$\sqrt{-1} = i$
$\sqrt{i} = \frac{1}{\sqrt{2}} \left(1 + i\right)$
$\sqrt{{\tilde \infty}} = {\tilde \infty}$
$\sqrt{\infty} = \infty$
$\sqrt{{e}^{i \theta} \infty} = {e}^{i \theta / 2} \infty$

$\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left[{z}^{2} - c\right] = \left\{\sqrt{c}, -\sqrt{c}\right\}$
$\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left[{z}^{2} - c\right] = \left\{i \sqrt{-c}, -i \sqrt{-c}\right\}$
$\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left[a {z}^{2} + b z + c\right] = \left\{\frac{-b + \sqrt{{b}^{2} - 4 a c}}{2 a}, \frac{-b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right\}$

## Functional equations

${\left(\sqrt{z}\right)}^{2} = z$
$\sqrt{{z}^{2}} = z$
$\sqrt{{x}^{2}} = \left|x\right|$
$\sqrt{-z} = i \sqrt{z}$
$\sqrt{-z} = -i \sqrt{z}$
$\sqrt{\frac{z}{2}} = \frac{\sqrt{z}}{\sqrt{2}}$
$\sqrt{a b} = \sqrt{a} \sqrt{b}$
$\sqrt{\frac{1}{z}} = \frac{1}{\sqrt{z}}$
$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{r {e}^{i \theta}} = \sqrt{r} {e}^{i \theta / 2}$
$\sqrt{z - c {z}^{2}} = \sqrt{z} \sqrt{1 - c z}$
$\sqrt{\frac{z}{z + c}} = \frac{\sqrt{z}}{\sqrt{z + c}}$
$\sqrt{\frac{z}{z - c}} = \frac{\sqrt{-z}}{\sqrt{c - z}}$
$\sqrt{\frac{z}{c - z}} = \sqrt{z} \sqrt{\frac{1}{c - z}}$

## Complex parts

$\left|\sqrt{z}\right| = \sqrt{\left|z\right|}$
$\arg\!\left(\sqrt{z}\right) = \frac{\arg(z)}{2}$
$\operatorname{sgn}\!\left(\sqrt{z}\right) = \sqrt{\operatorname{sgn}(z)}$
$\operatorname{Re}\!\left(\sqrt{z}\right) = \sqrt{\frac{\left|z\right| + \operatorname{Re}(z)}{2}}$
$\operatorname{Im}\!\left(\sqrt{z}\right) = \operatorname{sgn}\!\left(\operatorname{Im}(z)\right) \sqrt{\frac{\left|z\right| - \operatorname{Re}(z)}{2}}$
$\sqrt{\overline{z}} = \overline{\sqrt{z}}$

## Bounds and inequalities

$\left|\sqrt{z}\right| = \sqrt{\left|z\right|}$
$\left|\sqrt{x + a} - \sqrt{x}\right| \le \sqrt{x} \left(1 - \sqrt{1 - \frac{\left|a\right|}{x}}\right)$
$\left|\sqrt{x + a} - \sqrt{x}\right| \le \frac{\sqrt{x}}{2} \left(\frac{\left|a\right|}{x} + \frac{{\left|a\right|}^{2}}{{x}^{2}}\right)$
$\left|\frac{1}{\sqrt{x + a}} - \frac{1}{\sqrt{x}}\right| \le \frac{\left|a\right|}{2 {\left(x - \left|a\right|\right)}^{3 / 2}}$

## Derivatives and integrals

$\frac{d}{d z}\, \sqrt{z} = \frac{1}{2 \sqrt{z}}$
$\frac{d^{2}}{{d z}^{2}} \sqrt{z} = -\frac{1}{4 {z}^{3 / 2}}$
$\frac{d^{r}}{{d z}^{r}} \sqrt{z} = {\left(-1\right)}^{r} \left(-\frac{1}{2}\right)_{r} {z}^{r - 1 / 2}$
$\int_{a}^{b} \sqrt{z} \, dz = \frac{2}{3} \left({b}^{3 / 2} - {a}^{3 / 2}\right)$

## Series expansions

$\sqrt{z + x} = \sqrt{z} \sum_{k=0}^{\infty} \frac{{\left(-1\right)}^{k} \left(-\frac{1}{2}\right)_{k}}{{z}^{k} k !} {x}^{k}$
$\frac{1}{\sqrt{z + x}} = \frac{1}{\sqrt{z}} \sum_{k=0}^{\infty} \frac{{\left(-1\right)}^{k} \left(\frac{1}{2}\right)_{k}}{{z}^{k} k !} {x}^{k}$
${c}_{n} = \frac{1}{n {a}_{0}} \sum_{k=1}^{n} \left(\frac{3 k}{2} - n\right) {a}_{k} {c}_{n - k}\; \text{ where } {c}_{n} = [{x}^{n}] \sqrt{A},\;{a}_{n} = [{x}^{n}] A$
${c}_{n} = \frac{1}{n {a}_{0}} \sum_{k=1}^{n} \left(\frac{k}{2} - n\right) {a}_{k} {c}_{n - k}\; \text{ where } {c}_{n} = [{x}^{n}] \frac{1}{\sqrt{A}},\;{a}_{n} = [{x}^{n}] A$

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2021-03-15 19:12:00.328586 UTC