Question

If $\theta =\frac{\pi }{2}$ then the ${10}^{th}$ term of the series
$1+\left(\cos \theta +i\sin \theta \right)+{\left(\cos \theta +i\sin \theta \right)}^2+....is$

• A. $i$
• B. $-1$
• C. $1$
• D. $-i$

Answer 1

The correct option is A $i$

$1+(\cos \theta +i\sin \theta )+(\cos \theta +i\sin \theta {)}^2....$

So the tenth term = $(\cos \theta +i\sin \theta {)}^9$

Using de Moivre's theorm = $e^{i\theta 9}$

Putting $\theta =\frac{x}{2}$ , we get

$e^{i9\frac{x}{2}}$

$=e^{i\frac{x}{2}}$

$=\cos \frac{x}{2}+i\sin \frac{x}{2}$

$=i$