If #z+1/z=1#, find #z^3#, and then #z^1000+1/z^1000#?

2 Answers
Aug 9, 2018

Answer:

Please see the explanation below.

Explanation:

Let

#z=(costheta+isintheta)#

#1/z=(costheta-isintheta)#

#z+1/z=2costheta=1#

#costheta=1/2#

#theta=pi/3#, #[mod 2pi]#

Therefore,

#z=cos(pi/3)+isin(pi/3)#

And

#z^3=cos(3*pi/3)+isin(3*pi/3)=-1#

Avcoording to Demoivre' s Theorem

#z^1000=cos(1000*pi/3)+isin(1000*pi/3)#

#1/z^1000=cos(1000*pi/3)-isin(1000*pi/3)#

#z^1000+1/z^1000=2cos(1000/3pi)=-2*0.5=-1#

Aug 9, 2018

Answer:

#z^3 = -1" "# and #" "z^1000 + 1/z^1000 = -1#

Explanation:

Given:

#z+1/z = 1#

Note that:

#z^3+1 = (z+1)(z^2-z+1) = (z+1)z(z+1/z-1) = 0#

So:

#z^3 = -1#

So:

#z^1000+1/z^1000 = z^(3*333+1)+1/z^(3*333+1)#

#color(white)(z^1000+1/z^1000) = (-1)^333(z+1/z)#

#color(white)(z^1000+1/z^1000) = -(z+1/z)#

#color(white)(z^1000+1/z^1000) = -1#