Using de-moivre's theorem to evaluate Z^8, given that Z=1+i√3 ?

Question

2488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T17:39:29

The answer is #=128(-1+isqrt3)#

Convert the complex number to the trigonometric form

#z=a+ib#

#z=|z|*(costheta+isintheta)#

#|z|=sqrt(a^2+b^2)#

#costheta=a/(|z|)#

#sintheta=b/(|z|)#

Here,

#z=1+isqrt3#

#|z|=sqrt(1^2+(sqrt(3))^2)=sqrt4=2#

Therefore,

#z=2(1/2+isqrt3/2)#

#costheta=1/2#, #=>#, #theta=1/3pi#, #[mod 2pi]#

#z=2(cos(pi/3)+isin(pi/3))#

According to Demoivre's theorem,

#z=costheta+isintheta#

#z^n=(costheta+isintheta)^n=cos(ntheta)+isin(ntheta)#

Thetefore,

#z^8=(2(cos(pi/3)+isin(pi/3)))^8#

#=2^8(cos(8/3pi)+isin(8/3pi))#

#=256(-1/2+isqrt3/2)#

#=128(-1+isqrt3)#