Question 1134034
<br>
deMoivre's Theorem (unfortunately not in the curriculum in many high schools....):<br>
{{{(a(cos(b)+i*sin(b)))^n = (a^n)*(cos(n*b)+i*sin(n*b))}}}<br>
In your problem, a=16, b=200, n = 1/4, so<br>
{{{(16(cos(200)+i*sin(200)))^(1/4) = (16^(1/4))*(cos((1/4)*200)+i*sin((1/4)*200)) = 2*(cos(50)+i*sin(50))}}}<br>
A result of deMoivre's Theorem is that the n n-th roots of any complex number all have the same magnitude and are distributed around a circle in increments of 360/n degrees.  So the four 4th roots of your number are<br>
{{{2*(cos(50)+i*sin(50))}}}
{{{2*(cos(140)+i*sin(140))}}}
{{{2*(cos(230)+i*sin(230))}}}
{{{2*(cos(320)+i*sin(320))}}}<br>
Note we can use deMoivre's Theorem to verify that each of those is a 4th root of the given complex number.  For example,<br>
{{{(2*(cos(230)+i*sin(230)))^4 = (2^4)*(cos(4*230)+i*sin(4*230)) = 16*(cos(920)+i*sin(920)) = 16*(cos(200)+i*sin(200))}}}