Question

If $n$ is an integer and
$(1+i\sqrt{3}{)}^n+(1-i\sqrt{3}{)}^n=2^{n+1}\cos \theta$ then $\theta$=

• A. $\frac{n\pi }{3}$
• B. $\frac{n\pi }{2}$
• C. $\frac{n\pi }{4}$
• D. $\frac{n\pi }{6}$

Answer 1

Answer: (A)

$\frac{n\pi }{3}$

${(1+i\sqrt{3})}^n+{(1-i\sqrt{3})}^4=2^{n+1}\cos \theta$

$\Rightarrow 2^n{\left(e^{i\pi /3}\right)}^n+2^n{\left(e^{-1\pi /3}\right)}^n=2^{n+1}\cos \theta$

$\Rightarrow e^{in\pi /3}+e^{-in\pi /3}=2\cos \theta$

$\Rightarrow \cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}+\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3}=2\cos \theta$

$\Rightarrow 2\cos \frac{n\pi }{3}=2\cos \theta$

$\Rightarrow \frac{n\pi }{3}=\theta$

Hence, the answer is $\frac{n\pi }{3}.$