# taking square root algebraically

For getting the square root of the complex number^{} $a+ib$
($a,b\in \mathbb{R}$) purely algebraically, one should solve the real part^{} $x$ and the imaginary part $y$ of $\sqrt{a+ib}$ from the binomial equation

${(x+iy)}^{2}=a+ib.$ | (1) |

This gives

$$a+ib={x}^{2}+2ixy-{y}^{2}=({x}^{2}-{y}^{2})+i\cdot 2xy.$$ |

Comparing (see equality (http://planetmath.org/EqualityOfComplexNumbers)) the real parts and the imaginary parts yields the pair of real equations

$${x}^{2}-{y}^{2}=a,2xy=b,$$ |

which may be written

$${x}^{2}+(-{y}^{2})=a,{x}^{2}\cdot (-{y}^{2})=-\frac{{b}^{2}}{4}.$$ |

Note that the $x$ and $y$ must be chosen such that their product^{} ($=\frac{b}{2}$) has the same sign as $b$. Using the properties of quadratic equation, one infers that ${x}^{2}$ and $-{y}^{2}$ are the roots of the equation

$${t}^{2}-at-\frac{{b}^{2}}{4}=0.$$ |

The quadratic formula gives

$$t=\frac{a\pm \sqrt{{a}^{2}+{b}^{2}}}{2},$$ |

and since $-{y}^{2}$ is the smaller root, ${x}^{2}=\frac{a+\sqrt{{a}^{2}+{b}^{2}}}{2},-{y}^{2}=\frac{a-\sqrt{{a}^{2}+{b}^{2}}}{2}$. So we obtain the result

$$x=\sqrt{\frac{\sqrt{{a}^{2}+{b}^{2}}+a}{2}},y=(\mathrm{sign}b)\sqrt{\frac{\sqrt{{a}^{2}+{b}^{2}}-a}{2}}$$ |

(see the signum function). Because both may have also the

$\sqrt{a+ib}=\pm \left(\sqrt{{\displaystyle \frac{\sqrt{{a}^{2}+{b}^{2}}+a}{2}}}+(\mathrm{sign}b)i\sqrt{{\displaystyle \frac{\sqrt{{a}^{2}+{b}^{2}}-a}{2}}}\right).$ | (2) |

The result shows that the real and imaginary parts of the square root of any complex number $a+ib$ can be obtained from the real part $a$ and imaginary part $b$ of the number by using only algebraic operations, i.e. the rational operations^{} and the . Apparently, the same is true for all roots of a complex number with index (http://planetmath.org/NthRoot) an integer power of 2.

In practise, when determining the square root of a non-real complex number, one need not to remember the (2), but it’s better to solve concretely the equation (1).

Exercise. Compute $\sqrt{i}$ and check it!

Title | taking square root algebraically |

Canonical name | TakingSquareRootAlgebraically |

Date of creation | 2015-06-14 16:31:35 |

Last modified on | 2015-06-14 16:31:35 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 17 |

Author | pahio (2872) |

Entry type | Derivation |

Classification | msc 30-00 |

Classification | msc 12D99 |

Synonym | square root of complex number |

Related topic | SquareRootOfSquareRootBinomial |

Related topic | CasusIrreducibilis |

Related topic | TopicEntryOnComplexAnalysis |

Related topic | ValuesOfComplexCosine |