Question

If $a=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}$ then $(1+a{)}^{3n}$

• A. $-1$
• B. $0$
• C. $1$
• D. $(-1{)}^n$

Answer 1

The correct option is A $-1$

$a=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}=w^2$

Hence,

$(1+a{)}^{3n}=(1+w^2{)}^{3n}=(-w{)}^{3n}=-1(w^3{)}^n=(-1)1=-1$..............[since, $1+w+w^2=0$ and $w^3=1$ in general]