Find the real part of (sqrt{3} - i)^{2011}.

Answer 1

let's use De Moivre's Theorem

let z = √3 - 1
sketching it in the Argand plane will give you
|r| = 2, and tanØ = -1/√3
Ø = 330° or 11π/6

z = 2(cos 330° - i sin 330°)
z^2011 = 2^2011(cos (663630°) - i sin(663630°) )
reducing 663630° to a standard angle by dividing multiples of 360°
663630° -----> 150°

z^2011 = 2^2011 (cos 150 + i sin 150°)
= 2^(-√3/2 + (1/2) i )

so the real part is (2^2011)(-√3/2)
= - 2^2010 √3

Answer 2

clearly I have a typo in the 3rd last line, should be

= 2^2011 (-√3/2 + (1/2) i )