Question

Find the value of the expression:
$(\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2})(\cos \dfrac{\pi }{{{2^2}}} + i\sin \dfrac{\pi }{{{2^2}}})........\infty $

Answer 1

Hint: we know that the value of $(\cos \theta + i\sin \theta )$is ${e^{i\theta }}$. So, here the value of $(\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2})$would be ${e^{i\dfrac{\pi }{2}}}$ and so on. After combining all the exponential powers, we will get a geometric progression series (GP series) up to infinity. Sum of all the terms in an infinite GP is $\dfrac{a}{{1 - r}}$where $a$is the first term and $r$is the common ratio.

Complete Step by Step Solution:
$(\cos \theta + i\sin \theta ) = {e^{i\theta }}$
$ \Rightarrow (\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}) = {e^{i\dfrac{\pi }{2}}}$and $(\cos \dfrac{\pi }{{{2^2}}} + i\sin \dfrac{\pi }{{{2^2}}}) = {e^{i\dfrac{\pi }{{{2^2}}}}}$ and so on.
Writing the whole expression and its value;
Let $P = $$(\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2})(\cos \dfrac{\pi }{{{2^2}}} + i\sin \dfrac{\pi }{{{2^2}}})........\infty $
$P$= ${e^{i\dfrac{\pi }{2}}} \times {e^{i\dfrac{\pi }{{{2^2}}}}} \times {e^{i\dfrac{\pi }{{{2^3}}}}} \times {e^{i\dfrac{\pi }{{{2^4}}}}}.......\infty $$ \to $(equation $1$)
If two or more functions having the same base are multiplied together, then the powers of all the functions will be added.
Ex: ${x^2} \times {x^7} \times {x^y} = {x^{2 + 7 + y}}$
Equation $1$will be reduced to,
= ${e^{i(\dfrac{\pi }{2} + \dfrac{\pi }{{{2^2}}} + \dfrac{\pi }{{{2^3}}} + \dfrac{\pi }{{{2^4}}} + ........\infty )}}$$ \to $(equation $2$)
Let $y = \dfrac{\pi }{2} + \dfrac{\pi }{{{2^2}}} + \dfrac{\pi }{{{2^3}}} + \dfrac{\pi }{{{2^4}}} + .........\infty $
Equation $2$will reduce to,
$P$= ${e^{iy}}$$ \to $(equation $3$)
$y = \dfrac{\pi }{2} + \dfrac{\pi }{{{2^2}}} + \dfrac{\pi }{{{2^3}}} + \dfrac{\pi }{{{2^4}}} + .........\infty $ forms an infinite GP series whose first term is $\dfrac{\pi }{2}$and the common ratio is $\dfrac{{\dfrac{\pi }{{{2^2}}}}}{{\dfrac{\pi }{2}}} = \dfrac{1}{2}$ (common ratio = $\dfrac{{{\text{second term}}}}{{{\text{first term}}}}$)
Sum of infinite terms in a GP whose first term is $a$and common ratio is $r$.
$a = \dfrac{\pi }{2}$
$r = \dfrac{1}{2}$
$\therefore y = \dfrac{{\dfrac{\pi }{2}}}{{1 - \dfrac{1}{2}}}$$ = \dfrac{{\dfrac{\pi }{2}}}{{\dfrac{1}{2}}} = \pi $
Putting the value of $y$in equation $3$,
$P = {e^{i\pi }}$
Hence, the value of the expression $(\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2})(\cos \dfrac{\pi }{{{2^2}}} + i\sin \dfrac{\pi }{{{2^2}}})........\infty $would be ${e^{i\pi }}$.

Additional information:
As we know that the value of $(\cos \theta + i\sin \theta )$is ${e^{i\theta }}$. If we are asked to calculate the value of ${e^{ - i\theta }}$, we just need to substitute $ - \theta $in place of $\theta $.

So, the value of ${e^{ - i\theta }}$would be ($\cos \theta - i\sin \theta $).

Note:
The thing that should come to mind after seeing this question is that the value of $(\cos \theta + i\sin \theta )$is ${e^{i\theta }}$. In place of $\theta $, anything can be there. But the procedure will always be the same. Also remember the formula of infinite terms of GP.