Section 8.3 : De Moivre s Theorem and Applications


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1 The Sectio 8 : De Moivre s Theorem ad Applicatios Let z 1 ad z be complex umbers, where z 1 = r 1, z = r, arg(z 1 ) = θ 1, arg(z ) = θ z 1 = r 1 (cos θ 1 + i si θ 1 ) z = r (cos θ + i si θ ) ad z 1 z = r 1 r (cos θ 1 cos θ si θ 1 si θ } {{ } ) + i(si θ 1 cos θ + cos θ 1 si θ ) } {{ } cos(θ 1 +θ ) si(θ 1 +θ ) This meas or : = r 1 r (cos(θ 1 + θ ) + i si(θ 1 + θ )) 1 z 1 z = r 1 r = z 1 z arg(z 1 z ) = θ 1 + θ = arg(z 1 ) + arg(z ) The modulus of the product of two complex umbers is the product of their moduli, ad the argumet of the product of two complex umbers is the sum of their argumets We ca use these facts to compute the square of a complex umber (i polar form): suppose z = r(cos θ + i si θ), so z = r ad arg(z) = θ The z has modulus r r = r, ad z has argumet θ + θ = θ, ie z = r (cos(θ) + i si(θ)) This priciple ca be used to compute ay positive iteger power of z to give : Theorem 81: (De Moivre s Theorem) Let z = r(cos θ + i si θ), ad let be a positive iteger The z = r (cos θ + i si θ) 1
2 (ie i takig the th power of z, we raise the modulus to its th power ad multiply the argumet by ) Remark: Provided z 0, De Moivre s Theorem also holds for egative itegers We ow cosider three problems of differet types, all ivolvig De Moivre s theorem 1 Computig Positive Powers of a Complex Number Example 8 Let z = 1 i Fid z 10 Solutio: First write z i polar form z = 1 + ( 1) = arg(z) = π 4 (or 7π 4 ) Polar Form : z = ( cos( π 4 ) + i si( π 4 )) Applyig de Moivre s Theorem gives : z 10 = ( ( ) 10 cos(10 ( π ) 4 )) + i si(10 ( π 4 )) ( = cos( 10π ) 4 ) + i si( 10π 4 ) = (cos( π ) ) + i si( π ) = (cos( π ) + π) + i si( π + π) = (cos( π ) ) + i si( π ) = (0 + i ( 1)) = i Note: It ca be verified directly that (1 i) 10 = i Exercise : Use De Moivre s Theorem to fid (1 + i) 6 Computig th roots of a complex umber Example 8 Fid all complex cube roots of 7i Solutio: We are lookig for complex umbers z with the property z = 7i Strategy: First we write 7i i polar form : 7i = 0 + 7i = 0 + (7) = 7
3 arg(7i) = π 7i = 7(cos π + i si π ) Now suppose z = r(cos θ+i si θ) satisfies z = 7i The, by De Moivre s Theorem, r (cos θ + i si θ) = 7i = 7(cos π + i si π ) Thus r = 7 = r = (sice r must be a positive real umber with cube 7) What are the possible values of θ? We must have cos θ = cos π ad si θ = si π This meas : θ = π + πk, where k is a iteger; ie θ differs from π by a multiple of π Possibilities are : 1 k = 0: θ = π, θ = π 6 z 1 = (cos π 6 + i si π 6 ) = ( + i 1 ) z 1 = + i k = 1: θ = π + π(1) = π, θ = π 6 z = (cos π 6 + i si π 6 ) = ( + i 1 ) z = + i k = : θ = π + π() = 9π, θ = 9π 6 = π z = (cos π + i si π ) = (0 + i( 1)) z 1 = i
4 These are the oly possibilities : settig k = results i θ = π + π which gives the same result as k = 0 The complex cube roots of 7i are : z 1 = + i z = + i z = i z z z 1 I geeral : To fid the complex th roots of a ozero complex umber z 1 Write z i polar form : z = r(cos θ + i si θ) z will have differet th roots (ie cube roots, 4 fourth roots, etc) All these roots will have the same modulus r 1 (the positive real th root of r) 4 They will have differet argumets : θ, θ + π, θ + ( π) θ + (( 1) π),, The complex th roots of z are give (i polar form) by ( z 1 = r 1 cos( θ ) + i si( θ )) ( z = r 1 cos( θ+π θ+π ) + i si) ( z = r 1 cos( θ+4π ) + i si( θ+4π)), etc Example: Fid all the complex fourth roots of 16 Solutio: First write 16 i polar form Modulus : 16 Argumet : π 16 = 16(cos π + i si π) 4
5 Fourth roots of 16 all have modulus =, ad possibilities for the argumet are : π 4, π + π = π 4 4, π + 4π = π 4 4, π + π = 7π 4 4 Fourth roots of 16 are : z 1 = (cos( π 4 ) + i si( π 4 )) = + i z = (cos( π 4 ) + i si( π 4 )) = + i z = (cos( π 4 ) + i si( π 4 )) = i z 4 = (cos( 7π 4 ) + i si( 7π 4 )) = i Provig Trigoometric Idetities Example 84 : Prove that 1 cos θ = 16 cos θ 0 cos θ + cos θ si θ = 16 si θ 0 si θ + si θ Solutio: The idea is to write (cos θ + i si θ) i two differet ways We use both the biomial theorem ad De Moivre s theorem, ad compare the results Biomial Theorem: (cos θ + i si θ) = (cos θ) + (cos θ) 4 (i si θ) 1 + (cos θ) (i si θ) + (cos θ) (i si θ) 1 + (cos θ) 1 (i si θ) 4 + (cos θ) 0 (i si θ) 4 = cos θ + cos 4 θ(i si θ) + 10(cos θ)(i si θ) + 10(cos θ)(i si θ) +(cos θ)(i 4 si 4 θ) + (i si θ) = cos θ + i cos 4 θ si θ 10 cos θ si θ i 10 cos θ si θ + cos θ si 4 θ + i si θ = (cos θ 10 cos θ si θ + cos θ si 4 θ) + i( cos 4 θ si θ 10 cos θ si θ + si θ) Also, by De Moivre s Theorem, we have (cos θ + i si θ) = cos θ + i si θ
6 ad so cos θ + i si θ = (cos θ 10 cos θ si θ + cos θ si 4 θ) +i( cos 4 θ si θ 10 cos θ si θ + si θ) Equatig the real parts gives cos θ = cos θ 10 cos θ si θ + cos θ si 4 θ = cos θ 10 cos θ(1 cos θ) + cos θ(1 cos θ) = cos θ 10 cos θ + 10 cos θ + cos θ(1 cos θ + cos 4 θ) = cos θ 10 cos θ + 10 cos θ + cos θ 10 cos θ + cos θ cos θ = 16 cos θ 0 cos θ + cos θ For the other idetity, look at the imagiary parts : si θ = cos 4 θ si θ 10 cos θ si θ + si θ = (1 si θ) si θ 10(1 si θ) si θ + si θ = (1 si θ + si 4 θ) si θ 10 si θ + 10 si θ + si θ = si θ 10 si θ + si θ 10 si θ + 10 si θ + si θ = si θ 10 si θ + si θ 10 si θ + 10 si θ + si θ si θ = 16 si θ 0 si θ + si θ Remark: This method ca be used to prove may trigoometric idetities I geeral oe ca write si θ ad cos θ i terms of powers of si θ ad cos θ by usig both the biomial theorem ad De Moivre s theorem to expad (cos θ +i si θ) ad comparig the real ad imagiary parts of the results Exercise: Prove : 1 cos 4θ = 8 cos 4 θ 8 cos θ + 1 si 4θ = 4 cos θ si θ 4 cos θ si θ 6
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