taking square root algebraically
For getting the square root of the complex number () purely algebraically, one should solve the real part and the imaginary part of from the binomial equation
(1) |
This gives
Comparing (see equality (http://planetmath.org/EqualityOfComplexNumbers)) the real parts and the imaginary parts yields the pair of real equations
which may be written
Note that the and must be chosen such that their product () has the same sign as . Using the properties of quadratic equation, one infers that and are the roots of the equation
The quadratic formula gives
and since is the smaller root, . So we obtain the result
(see the signum function). Because both may have also the
(2) |
The result shows that the real and imaginary parts of the square root of any complex number can be obtained from the real part and imaginary part 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 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 |