Answer to Question #114657 in Linear Algebra for Struggling Student

Question

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

(a) Show that z (k) = ρ^1/6 [cos((π+2kπ)/6) + isin((π+2kπ)/6)], k = 0, 1, 2, 3, 4, 5 is a formula for the 6th roots of w. Show all your working.

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

(c) Given z = cosθ + isinθ and u + iv = (1 + z)(1 + z^2). Prove that v = utan(3θ/2) and u2 + v2 = 16cos^2(θ/2)cos^2(θ)

Answer 1

(a) Let $w=\rho(\cos\phi+i\sin\phi)$and $z=r(\cos\theta+i sin\theta)$ then by De Moivre's formula we have that $z^n=r^n(\cos n\theta+i \sin n\theta)$.

$z^n=w \rightsquigarrow r^n(\cos n\theta+i \sin n\theta)=\rho(\cos\phi+i\sin\phi)$

Hence, $r^n=\rho, \,\, n\theta=\phi+2\pi k$

$w\in\mathbb R_- \rightsquigarrow \phi=\pi$

For $n=6\colon r=\rho^{1/6} \bigg( \cos \frac{\pi+2\pi k}{6} +i\sin \frac{\pi+2\pi k}{6} \bigg)$

for each $k\in\{0,1,\dots,n-1=5\}$

(b) Let $w=-729 \rightsquigarrow \rho=729=3^6$

Hence, $\sqrt[6]{w}=3\bigg(\cos\frac{\pi+2\pi k}{6} +i\sin \frac{\pi+2\pi k}{6} \bigg), k\in\{0,1,\dots,5\}$

(c) $u+iv=(1+z)(1+z^2)=1+z+z^2+z^3=$

$=(1+\cos\theta+\cos2\theta+\cos3\theta)+i(\sin\theta+\sin2\theta+\sin3\theta)$

Hence, $u=1+\cos\theta+\cos2\theta+\cos3\theta$

$v=\sin\theta+\sin2\theta+\sin3\theta$, and

$\dfrac{v}{u}=\dfrac{\sin\theta+\sin2\theta+\sin3\theta}{1+\cos\theta+\cos2\theta+\cos3\theta}=\tan3\theta/2$ [1]

$u^2+v^2=$

$=(1+\cos\theta+\cos2\theta+\cos3\theta)^2+(\sin\theta+\sin2\theta+\sin3\theta)^2=$

$=4(\cos\theta/2+\cos3\theta/2)^2$ [2]

$=4\bigg( 2\cos \theta \cos\theta/2 \bigg)^2=16\cos^2 \theta \cos^2\theta/2$

[1] https://www.wolframalpha.com/input/?i=Simplify%5B%28sin+x%2Bsin+2x+%2Bsin+3x%29%2F%281%2Bcos+x%2Bcos+2x%2Bcos+3x%29%5D

[2] https://www.wolframalpha.com/input/?i=Simplify%5B%28sin+x%2Bsin+2x+%2Bsin+3x%29%5E2%2B%281%2Bcos+x%2Bcos+2x%2Bcos+3x%29%5E2%5D