Question 280701
DeMoivre's Theorem is usually used to raise complex numbers to a power. So I suspect that you left out an exponent. If so and if you are still not able to figure out the problem after reading what follows, please repost your question.<br>
(Note: For unknown reasons Algebra.com's formula software does not handle notation for inverse functions well. So I will use "acos" in place of cos^-1 and asin for sin^-1 in the expressions below.)
If all you have to do is rewrite the complex number in polar form then the formula is:
{{{a + bi = sqrt(a^2 + b^2)(cos(acos(a/(sqrt(a^2 + b^2)))) + i*sin(asin(b/(sqrt(a^2 + b^2)))))}}}
Using this on your complex number we get:
{{{sqrt(3) + 1i = sqrt((sqrt(3))^2 + 1^2)(cos(acos(sqrt(3)/sqrt((sqrt(3))^2 + 1^2)))) + i*sin(asin(1/sqrt((sqrt(3))^2 + 1^2))))}}}
Simplifying:
{{{sqrt(3+1)(cos(acos(sqrt(3)/sqrt(3+1))) + i*sin(asin(1/sqrt(3+1))))}}}
{{{sqrt(4)(cos(acos(sqrt(3)/sqrt(4))) + i*sin(asin(1/sqrt(4))))}}}
{{{2(cos(acos(sqrt(3)/2)) + i*sin(asin(1/2)))}}}
Using degrees, {{{acos(sqrt(3)/2) = asin(1/2) = 30}}}. (Use {{{pi/6}}} for radians.) We now have:
{{{2(cos(30) + i*sin(30))}}}
With polar form and DeMoivre's Theorem, it is very easy to raise complex numbers to powers.