Mathematics
De Moivre's Theorem
De Moivre's Theorem is a formula that relates complex numbers to trigonometry. It states that for any complex number z and any positive integer n, the nth power of z can be expressed in terms of its magnitude and argument. This theorem is useful in solving problems involving complex numbers and in understanding the behavior of periodic functions.
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4 Key excerpts on "De Moivre's Theorem"
eBook - PDF- Reg Allenby(Author)
- 1997(Publication Date)
- Butterworth-Heinemann(Publisher)
The important concepts of modulus and argument were introduced via polar representation of complex numbers, and their more prominent properties - in particular the triangle inequality - were established. The neat polar form of complex multiplication leads, naturally, to De Moivre's Theorem with its applications to finding roots and obtaining trigonometrical identities. Complex numbers, in particular the theory of functions of a complex variable, are important in many branches of mathematics, physics and engineering where complex functions are able to express, mathematically, the idea of flows and potentials. 176 and
eBook - PDF- Ron Larson(Author)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
Each of these numbers is a sixth root of 1. In general, an nth root of a complex number is defined as follows. Definition of an nth Root of a Complex Number The complex number u = a + bi is an nth root of the complex number z when z = u n = (a + bi) n . To find a formula for an nth root of a complex number, let u be an nth root of z, where u = s(cos β + i sin β) and z = r(cos θ + i sin θ ). By DeMoivre’s Theorem and the fact that u n = z, you have s n (cos nβ + i sin nβ) = r(cos θ + i sin θ ). Taking the absolute value of each side of this equation, it follows that s n = r. Substituting back into the previous equation and dividing by r gives cos nβ + i sin nβ = cos θ + i sin θ . So, it follows that cos nβ = cos θ and sin nβ = sin θ . Both sine and cosine have a period of 2π , so these last two equations have solutions if and only if the angles differ by a multiple of 2π . Consequently, there must exist an integer k such that nβ = θ + 2π k β = θ + 2π k n . Substituting this value of β and s = n radical.alt2r into the trigonometric form of u gives the result stated on the next page. Abraham DeMoivre (1667–1754) is remembered for his work in probability theory and DeMoivre’s Theorem. His book The Doctrine of Chances (published in 1718) includes the theory of recurring series and the theory of partial fractions. North Wind Picture Archives/Alamy Stock Photo Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
eBook - PDFMathematical Methods
A Course of Mathematics for Engineers and Scientists
- Brian H. Chirgwin, Charles Plumpton(Authors)
- 2014(Publication Date)
- Pergamon(Publisher)
434 A COURSE OF MATHEMATICS [CH. 7 15. The complex number z satisfies the equation (z-aY = b where a and b are complex numbers. Show that the points that represent the roots in the Argand plane are the vertices of a square and find the length of the side of this square in the case b = 1 + i. 7:5 Trigonometric expansions Two important applications of De Moivre's Theorem for integral n are : (i) the expansion of cos ηθ, sin nd in terms of cos 0 or sin 0, and (ii) the expression of cos 0, sin 0, in terms of cosines or sines of multiple angles. Expanding (cos Q + i sin θ) η by the binomial theorem and using De Moivre's Theorem, we find (cos ηθ + i sin ηθ) = (cos 0-fi sin θ) η = cos 0 - ( n j cos-2 0 sin 2 0 + y ) c o s -4 0 s i n 4 0 - . . . + i i / j ) cos'ho sin Θ -/ i c o s ^ e s i n 3 ^ . . . 1 . Equating real and imaginary parts gives cos ηθ and sin ηθ sepa-rately, viz. cos ηθ = cos 0 -( n J cos 2 0 sin 2 0-f ( n cos n ~ 4 0 sin 4 0- . . . , sin ηθ = / n cos-1 0 sin 0 -( cos~ 3 0 sin 3 0+ . . . . (7.22) It is not advisable to remember the formula for all the special cases which occur but to use the method indicated here in any particular problem. Use of the formula sin 2 0 = 1 — cos 2 0 enables cos ηθ to be expanded in powers of cos 0 for all n and in powers § 7 : 5 ] COMPLEX NUMBERS 435 of sin 0 (actually sin 2 0) when n is even. Similarly, when n is odd sin ηθ can be expressed in powers of sin 0 but, when n is even, sin ηθ/sin 0 can be expressed in powers of cos 0 only. Division of the second of eqns. (7.22) by the first gives (*) t a n ö - () tan 3 0 + ... tan ηθ = — ^ ^ . (7.23) 1 -Q ) t a n 2 0 + Q t a n * 0 - ... The formulae of eqns. (7.22) and (7.23) can be generalised by multiplying out the left-hand side of eqn. (7.4) and equating real and imaginary parts. Division then gives tan (0 1 + 0 2 + ...+θ η ) = -^———-^-—— (7.24) where t = tan 0i, etc. Example. (i) cos 30 = cos 3 0 -3 cos Θ sin 2 0 = 4 cos 3 0 -3 cos 0.
- K. F. Riley, M. P. Hobson(Authors)
- 2011(Publication Date)
- Cambridge University Press(Publisher)
The cube roots of unity are often written as 1, ω and ω 2 , with ω = e 2 iπ/ 3 . The properties ω 3 = 1 and 1 + ω + ω 2 = 0 are easily proved. 14 5.4.3 Solving polynomial equations A third application of de Moivre’s theorem is to the solution of general polynomial equations. The methods used are very similar to those employed when finding the roots of real polynomial equations. Indeed, the first step in solving complex polynomial equations is to use those same methods to obtain equations of reduced degree that are satisfied by z , or powers of z . The complex roots may then be deduced, as is illustrated in the following example. Example Solve the equation f ( z ) = z 6 − z 5 + 4 z 4 − 6 z 3 + 2 z 2 − 8 z + 8 = 0 . We first note that the sum of the coefficients in this sixth degree equation is zero; this means that z = 1 is one solution and that z − 1 is a factor of f ( z ). Either by inspection or by writing f ( z ) = ( z − 1)( z 5 + a 4 z 4 + a 3 z 3 + a 2 z 2 + a 1 z + a 0 ) and equating coefficients as in Section 2.1.1 , we find that f ( z ) = ( z − 1)( z 5 + 4 z 3 − 2 z 2 − 8) = 0 . The second term in parentheses can be factorised by inspection to give f ( z ) = ( z − 1)( z 3 − 2)( z 2 + 4) = 0 . Hence the roots are given by z 3 = 2, z 2 = − 4 and z = 1, with the solutions to the quadratic equation given immediately by z = ± 2 i . To find the complex cube roots, we first write the equation in the form z 3 = 2 = 2 e 2 ikπ , • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 14 This can be done either by direct substitution of the explicit forms or from the properties of the roots of a cubic equation as in Section 2.1.2 . 193 5.4 De Moivre’s theorem where k is any integer. If we now take the cube root of both sides, we get z = 2 1 / 3 e 2 ikπ/ 3 .
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