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
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 )
= 2^2011 (-√3/2 + (1/2) i )
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.