How do you use demoivre's theorem to simplify #2(sqrt3+i)^2#?

1 Answer
Sep 11, 2017

Answer:

#2(sqrt(3) + i)^2 = 4 + 4sqrt(3)i#

Explanation:

First convert your #z = sqrt(3) + i# into polar form by taking its absolute value and argument.

Recall that #|z| = sqrt((sqrt(3))^2 + 1^2) = 2#.

And that #Arg(z) = arctan(1/sqrt(3)) = pi/6#.

#:. z = 2* cis(pi/6)#.

By de Moivre's theorem, #z^n = r^n*cis(ntheta)#.

#:. z^2 = 4*cis(pi/3)#.

Converting back into Cartesian form, we have #x=r*cos(theta)# and #y=r*sin(theta)#

#:. x=4*cos(pi/3)# and #y=4*sin(pi/3)#.

#:. z^2=2+2sqrt(3)i#.

#:. 2z^2 = 4+4sqrt(3)i#.