Question 1180663
<br>
{{{-1+i*sqrt(3)}}} = {{{2cis(2pi/3)}}}<br>
{{{z^4 = 2cis(2pi/3)}}}<br>
To find the "first" solution using deMoivre's Theorem...<br>
{{{z = (2cis(2pi/3))^(1/4) = (2^(1/4))cis((2pi/3)/4) = 2^(1/4)cis(pi/6)}}}<br>
Then the other roots are spaced around the complex plane at increments of (2pi)/4=pi/2:<br>
{{{z=2^(1/4)cis(pi/6)}}}
{{{z=2^(1/4)cis(2pi/3)}}}
{{{z=2^(1/4)cis(7pi/6)}}}
{{{z=2^(1/4)cis(5pi/3)}}}<br>