Question

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 6
th root of w.
(a) Show that z (k) = ρ
1
6
-

cos

Answer 1

9.2 As per the given question,

As per demoiver's theorem,

$(\cos\theta+i\sin \theta)^5=\cos 5\theta +i\sin5\theta$

Now, let the binomial expansion of the left part,

$\cos^5\theta+5\cos^4\theta(i\sin\theta)+10\cos^3\theta(i\sin\theta)^2+10\cos^2\theta(i\sin\theta)^3+5\cos\theta(i\sin\theta)^4+(i\sin\theta)^5$

Now, equating the real and imaginary part of the above,

$\cos 5\theta =\cos^5\theta − 10 \cos^3 \theta \sin2\theta + 5 \cos\theta \sin4\theta$

Similarly,

$(\cos\theta+i\sin \theta)^4=\cos 4\theta +i\sin4\theta$

Now the expansion lhs of the equation,

$\cos^4\theta+4\cos^3\theta(i\sin\theta)+6\cos^2\theta(i\sin\theta)^2+4\cos\theta(i\sin\theta)^3+(i\sin\theta)^4$

Now, comparing the real and imaginary part,

$\sin 4\theta=4\cos^3\theta \sin \theta +4 \cos\theta \sin^3\theta$

9.3 As per the question, data is insufficient to answer 9.3, I am taking some assumptions,

that it is z is the complex number,

the polar form of the complex number, z= a($\cos\theta+i\sin \theta$ )

As per the question, it is given that w is the negative real number so,$(w)=(2k+1)\pi$

k=6,

$w=13\pi$

$z^6=a^6(\cos\theta+i\sin\theta)^6$

As per the demovier's rule

$=a^6(\cos6\theta+i\sin6\theta)$

6th real root =$\cos 6\theta=\cos 13\pi$ =-1

$z=re^{iθ}$ and convert1to polar form to get

$z^6=w$

z and w in polar form $z=re^{iθ},w=se^{iφ}$ . Then $z^6=w$

becomes:$(re^{iθ})^6=r^6e^{inθ}=se^{iφ}$

$r^6=s$

$e^{6\theta}=e^{i\phi}$

$r^6=s$

$r=s^{1/6}$

$θ=φ/n+2πl/n=2\pi/6=\pi/3+2πl/6$

6th root of the $=re^{i(\pi/3+2πl/6)}$

As data is not insufficient and incomplete so have chooses some approximation.