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[parent] taking square root algebraically (Derivation)

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
$\displaystyle a\!+\!ib = x^2\!+\!2ixy\!-\!y^2 =(x^2\!-\!y^2)\!+\!i\!\cdot\!2xy.$
Comparing (see equality) the real parts and the imaginary parts yields the pair of real equations
$\displaystyle x^2\!-\!y^2 = a,\quad 2xy = b,$
which may be written
$\displaystyle x^2\!+\!(-y^2) = a, \quad x^2\!\cdot\!(-y^2) = -\frac{b^2}{4}.$
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
$\displaystyle t^2\!-\!at\!-\!\frac{b^2}{4} = 0.$
The quadratic formula gives
$\displaystyle t = \frac{a\pm\sqrt{a^2\!+\!b^2}}{2},$
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
$\displaystyle x = \sqrt{\frac{\sqrt{a^2\!+\!b^2}+\!a}{2}}, \quad y = ($sign$\displaystyle \,b)\sqrt{\frac{\sqrt{a^2\!+\!b^2}-\!a}{2}}$
(see the signum function). Because both may have also the opposite signs, we have the formula
$\displaystyle \sqrt{a\!+\!ib} = \pm\left(\sqrt{\frac{\sqrt{a^2\!+\!b^2}+\!a}{2}}+(\mbox{sign}\,{b})i\sqrt{\frac{\sqrt{a^2\!+\!b^2}-\!a}{2}}\right).$ (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!



"taking square root algebraically" is owned by pahio.
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See Also: square root of square root binomial, casus irreducibilis, topic entry on complex analysis, values of complex cosine

Other names:  square root of complex number

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quadratic equation in $\mathbb{C}$ (Theorem) by pahio
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Cross-references: power, integer, rational, operations, algebraic, signum function, quadratic formula, roots, properties of quadratic equation, product, equations, real, binomial equation, imaginary part, real part, complex number, square root
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This is version 11 of taking square root algebraically, born on 2005-06-20, modified 2008-01-29.
Object id is 7175, canonical name is TakingSquareRootAlgebraically.
Accessed 3270 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 30-00 (Functions of a complex variable :: General reference works )
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