Question #113413
QUESTION 9

9.1 State De Moivre’s Theorem.
(2)

9.2 Express cos 5θ and sin 4θ as polynomials in terms of sin θ and cos θ.
(8)

9.3 Let w be a negative real number, z a 6th root of w.

(a) Show that z (k) = ρ 6

6th roots of w.

cos

( π+2kπ )

+ i sin

( π+2kπ )

, k = 0, 1, 2, 3, 4, 5 is a formula for the



Show all your working.
(8)

(b) Hence determine the 6th roots of −729.
(2)

(c) Given z = cos θ + i sin θ and u + iv = (1 + z)(1 + z2). Prove that v = u tan( 3θ ) and

u2 + v2 = 16 cos2( θ ) cos2(θ)
(10)

[30]
1
Expert's answer
2020-05-04T20:09:56-0400

9.1 The Moivre’s Theorem

z=r(cosθ+isinθ)zn=rn(cosnθ+isinnθ),nNz=r(\cos\theta+i\sin \theta)\\ z^n=r^n(\cos n\theta+i\sin n\theta), n\in N


9.2 Let r=1r=1

z=cosθ+isinθz5=cos5θ+isin5θ=(cosθ+isinθ)5==cos5θ+5icos4θsinθ+10i2cos3θsin2θ++10i3cos2θsin3θ+5i4cosθsin4θ+i5sin5θ==cos5θ10cos3θsin2θ+5cosθsin4θ++i(5cos4θsinθ10cos2θsin3θ+sin5θ)cos5θ=cos5θ10cos3θsin2θ+5cosθsin4θz=\cos\theta+i\sin\theta\\ z^5=\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5=\\ =\cos^5\theta+5i\cos^4\theta\sin\theta+10i^2\cos^3\theta\sin^2\theta+\\ +10i^3\cos^2\theta\sin^3\theta+5i^4\cos\theta\sin^4\theta+i^5\sin^5\theta=\\ =\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta+\\ +i(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)\\ \cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\


z4=cos4θ+isin4θ=(cosθ+isinθ)4==cos4θ+4icos3θsinθ+6i2cos2θsin2θ++6i3cosθsin3θ+i4sin4θ==cos4θ6cos2θsin2θ+sin4θ++i(4cos3θsinθ6cosθsin3θ)sin4θ=4cos3θsinθ6cosθsin3θz^4=\cos4\theta+i\sin4\theta=(\cos\theta+i\sin\theta)^4=\\ =\cos^4\theta+4i\cos^3\theta\sin\theta+6i^2\cos^2\theta\sin^2\theta+\\ +6i^3\cos\theta\sin^3\theta+i^4\sin^4\theta=\\ =\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta+\\ +i(4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta)\\ \sin4\theta=4\cos^3\theta\sin\theta-6\cos\theta\sin^3\theta


i2=1,i3=i,i4=1,i5=ii^2=-1, i^3=-i, i^4=1,i^5=i


9.3 a)

z=w6,w<0w=ρ(cosθ+isinθ)z=ρ6(cosθ+2πk6+isinθ+2πk6),k=0,1,2,3,4,5z=\sqrt[6]{w}, w<0\\ w=\rho(\cos\theta+i\sin\theta)\\ z=\sqrt[6]\rho(\cos\frac{\theta+2\pi k}{6}+i\sin\frac{\theta+2\pi k}{6}), k=0,1,2,3,4,5


b)

7296=1366=3161=cosπ+isinπ16=cosπ+2πk6+sinπ+2πk6,k=0,1,2,3,4,57296=3(cosπ+2πk6+sinπ+2πk6),k=0,1,2,3,4,5\sqrt[6]{-729}=\sqrt[6]{-1\cdot3^6}=3\sqrt[6]{-1}\\ -1=\cos\pi+i\sin\pi\\ \sqrt[6]{-1}=\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}, k=0,1,2,3,4,5\\ \sqrt[6]{-729}=3(\cos\frac{\pi+2\pi k}{6}+\sin\frac{\pi+2\pi k}{6}),\\ k=0,1,2,3,4,5\\


7296=3(cosπ6+sinπ6)=3(32+i12),7296=3(cos3π6+sin3π6)=3i,7296=3(cos5π6+sin5π6)=3(32+i12),7296=3(cos7π6+sin7π6)=3(32i12),7296=3(cos9π6+sin9π6)=3i,7296=3(cos11π6+sin11π6)=3(32i12)\sqrt[6]{-729}=3(\cos\frac{\pi}{6}+\sin\frac{\pi}{6})=3(\frac{\sqrt3}{2}+i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{3\pi }{6}+\sin\frac{3\pi }{6})=3i,\\ \sqrt[6]{-729}=3(\cos\frac{5\pi }{6}+\sin\frac{5\pi }{6})=3(-\frac{\sqrt3}{2}+i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{7\pi}{6}+\sin\frac{7\pi}{6})=3(-\frac{\sqrt3}{2}-i\frac{1}{2}),\\ \sqrt[6]{-729}=3(\cos\frac{9\pi}{6}+\sin\frac{9\pi}{6})=-3i,\\ \sqrt[6]{-729}=3(\cos\frac{11\pi }{6}+\sin\frac{11\pi }{6})=3(\frac{\sqrt3}{2}-i\frac{1}{2})\\


c)

z=cosθ+isinθu+iv=(1+z)(1+z2)==((1+cosθ)+isinθ)(1+cos2θ+2icosθsinθsin2θ)==((1+cosθ)+isinθ)((1+cos2θsin2θ)+2icosθsinθ)==((1+cosθ)+isinθ)(2cos2θ+2icosθsinθ)==2cos2θ(1+cosθ)2cosθsin2θ++i(2cosθsinθ(1+cosθ)+2sinθcos2θ)u=2cos2θ(1+cosθ)2cosθsin2θ==2cosθ(cosθ+cos2θ)v=2cosθsinθ(1+cosθ)+2sinθcos2θ==2sinθcosθ(1+2cosθ)=2cosθ(sinθ+sin2θ)utan3θ2=2cosθ(cosθ+cos2θ)sin3θ2cos3θ2==2cosθcos3θ2sinθ2+sin5θ2sinθ2+sin7θ22==2cosθcos3θ22sin3θcosθ22==2cosθ(2sin3θ2cosθ2)==2cosθ(sinθ+sin2θ)=vz=\cos\theta+i\sin\theta\\ u+iv=(1+z)(1+z^2)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(1+\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\((1+\cos^2\theta-\sin^2\theta)+2i\cos\theta\sin\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(2\cos^2\theta+2i\cos\theta\sin\theta)=\\ =2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta+\\ +i(2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta)\\ u=2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta=\\ =2\cos\theta(\cos\theta+\cos2\theta)\\ v=2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta=\\ =2\sin\theta\cos\theta(1+2\cos\theta)=2\cos\theta(\sin\theta+\sin2\theta)\\ u\tan\frac{3\theta}{2}=2\cos\theta(\cos\theta+\cos2\theta)\frac{\sin\frac{3\theta}{2}}{\cos\frac{3\theta}{2}}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{\sin\frac{\theta}{2}+\sin\frac{5\theta}{2}-\sin\frac{\theta}{2}+\sin\frac{7\theta}{2}}{2}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{2\sin3\theta\cos\frac{\theta}{2}}{2}=\\ =2\cos\theta(2\sin\frac{3\theta}{2}\cos\frac{\theta}{2})=\\ =2\cos\theta(\sin\theta+\sin2\theta)=v


u2+v2=u=4cos2θ(cosθ+cos2θ)2++4cos2θ(sinθ+sin2θ)2==4cos2θ(cos2θ+2cosθcos2θ+cos22θ++sin2θ+2sinθsin2θ+sin22θ)==4cos2θ(2+2cosθ)=8cos2θ(1+cosθ)==16cos2θcos2θ2u^2+v^2=u=4\cos^2\theta(\cos\theta+\cos2\theta)^2+\\ +4\cos^2\theta(\sin\theta+\sin2\theta)^2=\\ =4\cos^2\theta(\cos^2\theta+2\cos\theta\cos2\theta+\cos^22\theta+\\ +\sin^2\theta+2\sin\theta\sin2\theta+\sin^22\theta)=\\ =4\cos^2\theta(2+2\cos\theta)=8\cos^2\theta(1+\cos\theta)=\\ =16\cos^2\theta\cos^2\frac{\theta}{2}



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