Question

Value of $i^{1947}+i^{(-1947)}$ is

• A. 2
• B. 0
• C.
• D. - 2i

Answer 1

The correct option is B

0

$\begin{array}{c}486\\ 41947\\ \underset{–}{16}2\\ 34\\ \underset{–}{32}\\ 27\\ \underset{–}{24}\\ 3\end{array}$

$i^{1947}$ = $i^{\left(\right(4\times 486)+3)}$ = ($i^4{)}^{486}$ × $i^3$

= $i^3$ = -i

$i^{(-1947)}$ = $\frac{1}{i^3}$ = $\frac{1}{(-i)}$ = i

$\therefore$ Answer is $-i+i$=0

Short cut: consider only last 2 digits of exponent.

The remainder of this number when divided by 4 will be the exponent

ie, $i^{1947}$ = $i^R$ : R=remander of$\left(\frac{47}{4}\right)$=3

= $i^3$ = -i

$i^{(-1947)}$ = $\frac{1}{i^3}$ = $\frac{1}{(-i)}$ = i

$\therefore$ Answer is $-i+i$=0