# 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

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

This gives

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

Comparing (see equality (http://planetmath.org/EqualityOfComplexNumbers)) 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 $x$ 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

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