Question #1385b

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Question #1385b

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Answer 1 · 2017-11-06T04:18:12

The answer is 1.

Explanation:

Since we know $i = \sqrt{- 1}$ $i = sqrt(-1)$ , we can say $i^{2} = - 1$ $i^2 = -1$ . By this logic, we can separate $i^{56}$ $i^56$ into ${(i^{2})}^{28}$ $(i^2)^28$ , which simplifies into ${(- 1)}^{28}$ $(-1)^28$ , or simply $1$ $1$ .

Answer 2 · 2017-11-06T08:44:34

$1$ $1$

Explanation:

remember that

$i = \sqrt{- 1}, i^{2} = - 1, i^{3} = - i, i^{4} = 1$ $i=sqrt(-1), i^2=-1,i^3=-i,i^4=1$

so we look for multiples of $4$ $4$ in order to simplify the problem simplify the problem

$i^{56} = i^{4 \times 14}$ $i^(56)=i^(4xx14)$

$= {(i^{4})}^{56}$ $=(i^4)^56$

$= 1^{56} = 1$ $=1^56=1$

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Question #1385b

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