Question

$x_r=\cos \frac{\pi }{2^r}+i\sin \frac{\pi }{2^r}$ then $x_1.x_2.x_3......\infty =-1$

• A. True
• B. False

Answer 1

The correct option is A True

We have,

$x_r=\cos \frac{\pi }{2^r}+i\sin \frac{\pi }{2^r}$

$x_r=e^{\frac{i\pi }{2^r}}$

Now,

$x_1.x_2.x_3......\infty$

$\Rightarrow e^{\frac{i\pi }{2^1}}.e^{\frac{i\pi }{2^2}}.e^{\frac{i\pi }{2^3}}......\infty$

$\Rightarrow e^{i\pi (\frac{1}{2}+\frac{1}{4}+\frac{1}{8}......\infty )}$

$\Rightarrow e^{i\pi \left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)}$

$\Rightarrow e^{i\pi \left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)}$

$\Rightarrow e^{i\pi \times 1}$

$\Rightarrow e^{i\pi }$

$\Rightarrow \cos \pi +i\sin \pi =-1$

Hence, this is the answer.