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🔎 The root of a complex number


A square root of a complex number z is a complex number you check you2=z. Every complex number has exactly two distinct square roots, except 0, for which 0 is the only square root. For example, the two square roots of -1 are i and –i Or i is thereimaginary unit (In mathematics, the imaginary unit is a complex number, denoted i, which has…). More generally, a nth root in z is a complex numbers (Complex numbers form an extension of the set of real numbers. They allow…) you check youn=z. Apart from 0, All (The whole understood as everything that exists is often interpreted as the world or…) no (The notion of number in linguistics is treated in the article “Number…) complex admitted exactly n root ndifferent -ths. For example, is a third root of -1.

root n– small in unit 1 form a group for the product, denoted Unwhich is a cyclic group (In mathematics and more precisely in algebra, a cyclic group, or what…) in order n.

There is no continued determination of a square root (The square root of a positive real number x is the positive number whose…) on C. More precisely, there are no such continuous applications f(z)2=z.

The formulas

Square root in Cartesian coordinates.
root k– that’s it polar coordinates (Polar coordinates are, in mathematics, a coordinate system…).

In polar coordinates, a complex number is written z = reii Or r is positive. Whether z is non-zero, r is positive. root k– that’s it z are complex numbers you like youk=z. They are in between k and is clearly provided by:

Or .

For k=2, we get a square root description. Verification is an application of Moivre Formula (De Moivre’s formula (in reference to Abraham de Moivre) or de Moivre’s formula (see article…).

Unit root

12th unit root, positioned on the dry (A circle is a closed plane curve made up of equal points…) unit.

root k-ths in unit 1 are sometimes called there poor name. Product of a root n-th in 1 and a root k-th in 1 is a root r-th in 1, place r = pgcd(k,n). In particular, thereensemble (In set theory, a set intuitively designates a collection…) Uk root k-ths of 1 is stable not multiplication (Multiplication is one of the four operations in elementary arithmetic…) complex. It is a group finished (In mathematics, a finite group is a group that contains a finite number…). root k-Ti in 1 is written:

Or .

The group Uk is cyclic and its generator is called primitive root in 1.

All k-th unit roots lie on the unit circle and are the vertex regular polygon (In geometry, a regular polygon is an equilateral polygon (all of it…) and k where there is a vertex with the affix 1.


The above formulas demonstrate the existence of square roots and roots k-ths in the unit. But the formulas for the roots k-ths count on there definition (A definition is a statement of what something is or what a noun means. Hence…) in’exponential (The exponential function is one of the most important applications of analysis, or more…) complex and on the de Moivre formula. This implementation is defined as an array of integers and is an instance of an integer function. It is implicitly appealed here to the complex analysis.

There theorem (A theorem is a proposition that can be demonstrated mathematically, that is, a…) fundamental ofalgebra (Algebra, a word of Arabic origin al-jabr (الجبر), is the branch…)

Carl Friedrich Gauss (Johann Carl Friedrich Gauß (traditionally transcribed Gauss in French)…) to whom we owe the first solid proof of the fundamental theorem

The fundamental theorem of algebra can be applied, and it shows that any nonzero complex number has exactly k different complex roots.

The fundamental theorem of algebra has many proofs. Some are detailed in the dedicated article. Most use tools topology (Topology is a branch of mathematics concerned with the study of spatial deformation by…) and analysis, even complex analysis. Apply it here to avoid the formula de Moivre has a meaning (SENS (Strategies for Engineered Negligible Senescence) is a scientific project aimed at…) only if we have a proof that uses algebra alone. Like a demonstration (In mathematics, a demonstration makes it possible to establish a proposition from…) was published by the collective Nicolas Bourbaki (Nicolas Bourbaki is an imaginary mathematician, under his name a group…) in 1952, corrected an idea presented by Laplace in 1795.

Using the evidence attributed to d’Alembert and Argand would not be valid here. This proof, which was actually completed in 1941 by Littlewood, is based on a construction of the roots k-ths two complex numbers. This construction should not use the de Moivre formula.

Lemme de Littlewood

The proof of the fundamental theorem of algebra was proposed by Littlewood in 1941[réf. souhaitée] that’s essentially it data (In information technology (IT), data is an elementary description, often…) by d’Alembert and corrected by Argand and Cauchy, except that in the lemma the existence of at least one root k-th for every complex number, we avoid using the Moivre formula: we rely on the definition of the roots R+ and we use a bit of topology.

  • We first show that if k East disabledany complex z nonzero has at least one root k-th. The method (analogous to that of Cauchy for the proof of the fundamental theorem) is set K(X) = Xkzthey choose a complex you achieve the minimum | K(you) | on complex plan (In mathematics, the complex plane (also called the Cauchy plane) designates a plane where every…)and show that K(you) = 0.
    • We are starting to show that you is non-zero. For this, we put u0 = z / | z | (which is a complex number modulo 1), we choose u among 1, -1, i and -i so that | uu0 | , and we put v=uk. So (since u4= 1 and that k2 is congruent to 1 module (In modular arithmetic, we talk about congruent numbers modulo n term modulo can also be…) 4). Then we set. Then |Q(y)|=||z|uzYou are the true name of zero.
    • Then we deduce that K(you) = 0. For this, we write K(you + h) = (youkz) + kyouk – 1h + h2R(h) for a certain polynomial (A polynomial, in mathematics, is the linear combination of the products of…) R. For , we have where α is the sum of the modulus of the coefficients in R. For and 0so that , where (passes in the time limit t tends to 0) K(you) = 0.
  • Then we extend the previous result to the whole k> 0, by induction on its 2-value, i.e. on the biggest integer r like k must be divisible by 2r.
    • The case r=0 matches the case k odd deja vu.
    • The repetition step is done by noticing that if k=2q and yes you is one of the two square roots of z (given formula in Cartesian coordinates), any root q-th of you is the root k-th of z.

Once this lemma is established (in the existence of at least one root k-th), the proof of the fundamental theorem of algebra continues in the same way data (In information technology, a data is an elementary description,…) by d’Alembert and corrected by Argand and Cauchy. Then we deduce from this theorem (cf supra) that every nonzero complex number admits exactly k root k-ièmes complex.

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