Question 421149
{{{z^6 = 12+5i}}}
{{{z^6 = 13(12/13 + 5i/13)}}}
==> {{{z = root(6, 13)(12/13 + 5i/13)^(1/6)}}}
Let {{{theta}}} be the angle in quadrant 1 such that {{{cos (theta) = 12/13}}} and {{{sin(theta) = 5/13}}}, or {{{tan (theta) = 5/12}}}.  Then

{{{z = root(6, 13)(cos(theta) + i*sin(theta))^(1/6)}}}
<==> {{{z = root(6, 13)(cos(theta/6) + i*sin(theta/6))}}} by de Moivre's Theorem.

This is the principal 6th root.  The other roots are given by 
{{{z = root(6, 13)(cos(theta/6  +(pi*n)/3) + i*sin(theta/6 + (pi*n)/3))}}}, letting n = 1,2,3,4,5.