How do you find the fifth roots of #128(-1+i)#?

Question

3481 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T05:55:55

#color(purple)(=> (sqrt8 * (cos 27 + i sin 27)#

#128(-1+i) = 128sqrt2(-1/sqrt2 + i (1/sqrt2))#

#=> 128sqrt2 (-cos (pi/4) + i sin (pi/4))#

#=> 128 sqrt2 (cos (135) + i sin (135))#

#root(5) (128sqrt2 (cos 135 + i sin 135)#

#=> (128sqrt2)^(1/5) (cos (135/5) + i sin (135/5))#

#=> (sqrt8 * (cos 27 + i sin 27)# as

#(128 sqrt2)^(1/5) = (2^7 * sqrt 2)^(1/5) = (sqrt2)^(15/5) = sqrt8#