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!