Question

For $n\in N$, If $A_n=\cos \left(\frac{\pi }{2^n}\right)+i\sin \left(\frac{\pi }{2^n}\right)$, then ${\left(A_1{A}_2{A}_3{A}_4\right)}^4=$

• A. $\frac{-1-i}{\sqrt{2}}$
• B. $1$
• C. $0$
• D. $\frac{1-i}{\sqrt{2}}$

Answer 1

Answer: (D)

It is given that,
$A_n=\cos \left(\frac{\pi }{2^n}\right)+i\sin \left(\frac{\pi }{2^n}\right),n\in N=e^{i\left(\frac{\pi }{2^n}\right)}$
$\therefore {\left(A_1{A}_2{A}_3{A}_4\right)}^4=e^{i\pi {\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}^4}$
$=e^{i\frac{15\pi }{4}}=e^{i\left(4\pi -\frac{\pi }{4}\right)}$
$=\cos \left(4\pi -\frac{\pi }{4}\right)+i\sin \left(4\pi -\frac{\pi }{4}\right)$
$=\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}=\frac{1-i}{\sqrt{2}}$