Asked by Bob

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

Answers

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

= 2^2011 (-√3/2 + (1/2) i )
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