#2(sqrt(3) + i)^2 = 4 + 4sqrt(3)i#
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#.
How do you use demoivre's theorem to simplify #2(sqrt3+i)^2#?
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