Question 1170521
Noting that a {{{2pi}}} rotation doesn't change things, we can express z in polar form:
{{{ matrix(3,3,"","","","", z = 16e^(((4pi)/3 + 2pi*n)*i), "","","","") }}}
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Now finding the four 4th roots is nearly trivial (4th root of 16 is 2, then just divide exponents by 4).  You must let n=0,1,2, and 3 to get the four  4th roots:<br>
n=0:  {{{ z[0] = matrix(3,3,"","","","", 2e^(((pi)/3)*i) = 2(cos(pi/3) + i*sin(pi/3)), "","","","") }}}  = 2(0.5 + i*0.866) = {{{highlight(1 + i*1.732) }}}
(or {{{highlight(1+i*sqrt(3)) }}} if exact answer is needed)
NOTE that {{{z[0]}}} is also called the principal root <br>

Now set n=1 to find the 2nd root

n=1:  {{{ z[1] = matrix(3,3,"","","","", 2e^((((4pi)/3 + 2pi*1)/4) *i), "","","","") }}}

That exponent works out to {{{ 5pi/6 }}}:


{{{ z[1] = matrix(3,3,"","","","", 2e^(( (5pi)/6)*i) = 2(cos((5pi)/6) + i*sin((5pi)/6)), "","","","") }}} = {{{ highlight(-1.732 + i*1) }}}



Let n=2 then n=3 to get the remaining two roots.  You should get:
n=2:  {{{ z[2] = 2*(cos((4pi)/3) + i*sin((4pi)/3))) }}}
n=3:  {{{ z[3] = 2*(cos((11pi)/6) + i*sin((11pi)/6))) }}}