Answer to Question #114823 in Complex Analysis for Ntokozo

Question

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 (pi+2kpi/6 )+ isin ( pi+2kpi/6), , k = 0, 1, 2, 3, 4, 5 is a formula for the
6th roots of w.

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