Question #f6317

Question

3414 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-18T04:21:55

I have corrected the question. #(sqrt 3 - i)^3=-8i#

De Moivre's Theorem; If #cis theta = cos theta + i sin theta#, then

#(cis theta)^n = cis ntheta = cos ntheta + i sin ntheta#

compare #(sqrt 3 - i )# with #r cis theta#.

#r cos theta = sqrt 3 and r sin theta = -1#. Solving,

#r = 2 and theta = -pi/6#.

So, #(sqrt 3 - i )^3= (2(cis (-pi/6))^3= 2^3 cis (3(-pi/6))=8 cis (-pi/2)=8 (cos(- pi/2)+isin(-pi/2))=8(0-i)=-8i#.

.