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

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

This gives

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

Comparing (see equality) the real parts and the imaginary parts yields the pair of real equations

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

which may be written

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

Note that the signs of $x$ 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, $x^{2}=\frac{a\!+\!\sqrt{a^{2}\!+\!b^{2}}}{2},\quad-y^{2}=\frac{a\!-\!\sqrt{a^{% 2}\!+\!b^{2}}}{2}$. So we obtain the result

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

(see the signum function). Because both may have also the opposite signs, we have the formula

$\displaystyle\sqrt{a\!+\!ib}\;=\;\pm\left(\sqrt{\frac{\sqrt{a^{2}\!+\!b^{2}}+% \!a}{2}}+(\mathrm{sign}\,{b})i\sqrt{\frac{\sqrt{a^{2}\!+\!b^{2}}-\!a}{2}}% \right).$ | (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 radicals. Apparently, the same is true for all roots of a complex number with index an integer power of 2.

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

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

## Comments

## Archimedes' square root extimated irrational roots by three step

taking the square root of any rational number follows an older inverse proportion documented by Archimedes, Fibonacci and Galileo.

## Wanted: volunteer to make following entry

Would be glad if some member would volunteer to make the following entry in maths encyclopedia: Mangammal primes: These are the impossible prime factors of (3^n - 2). see A 123239 in OEIS. This is a corollary of Ëuler’s generalisation of Fermat’s theore- a further generalisation (ISSN 1550-3747 ).

## Wanted: volunteer to make following entry

Would be glad if some member would volunteer to make the following entry in maths encyclopedia: Mangammal primes: These are the impossible prime factors of (3^n - 2). see A 123239 in OEIS. This is a corollary of Ëuler’s generalisation of Fermat’s theore- a further generalisation (ISSN 1550-3747 ).

## Wanted: volunteer to make following entry

Would be glad if some member would volunteer to make the following entry in maths encyclopedia: Mangammal primes: These are the impossible prime factors of (3^n - 2). see A 123239 in OEIS. This is a corollary of Ëuler’s generalisation of Fermat’s theore- a further generalisation (ISSN 1550-3747 ).

## Wanted: volunteer to make following entry

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## Wanted: volunteer to make following entry

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## Using pari to find valid bases for pseudoprimes

Supposing we wish to find the base for which 77 is a pseudoprime we can use pari. The programme: p(n) =(n^76-1)/77 for(n=1, 60,print (p(n))). Thus I found that 34 is a valid base for pseudoprimality of 77.

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## Using pari to find valid bases for pseudoprimes(contd)

Not only 29, but also 41, 43, 62, 64, 71, 76, 83, 92, 97 and 104 are valid bases for pseudoprimality of 105.

## Using pari to find valid bases for pseudoprimes(contd)

In Z(i) (20 + 21*i) and (21 + 20*i) are valid bases for pseudoprimality of 105. Needless to say their conjugates are also valid bases.

## Using pari to find valid bases for pseudoprimes(contd)

I was looking for something predictable in finding valid bases for pseudoprimality of 105. Happy to say that I succeeded: starting from 41, 41 + 21*k, where k belongs to N are valid bases; exceptions - integers ending with 0 or 5. Needless to say there are bases other than these.

## How to find valid bases for pseudoprimes in Z(i)

We can use pari. There is another way: Take a known base for pseudoprimality of a composite number. Split it into two parts such that one part is exactly divisible by one or more prime factors of the given composite number and the other is exactly divisible by the remaining prime factor/s. Let any one of the two parts be the real and the other be the coefficient of i in the complex base. Example: 29 is a base for pseudo -primality of 105 in Z. 29 can be split into two parts 14 and 15. 14 is divisible by 7 and 15 is divisible by 3 and 5. Hence 14 + 15i and 15 + 14i are valid bases in Z(i). Needless to say conjugates of these two are also valid bases.