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3 votes
$i^5+i^{-25}+i^{45}$
$i^5+i^(-25)+i^(45)$

2 Answers

2 votes

Answer:

1/i +2i

Explanation:

i^5+i^-25+i^45 i^2=-1

i^4*i +i^-24*i^-1 +i^44 *i

(i²)² i+ (i²)^-12*i^-1+(i²)^22 *i since i²=1

i+i^-1+i=

i+1/i +i=1/i +2i

User Ekos
by
4.6k points
7 votes


i^5+i^(-25)+i^(45)

Rewrite the term
i^(-25)


=i^5+(1)/(i^(25))+i^(45)

Expand each term so we have


=i(i^2)^2+(1)/(i(i^2)^(12))+i(i^2)^(22)

Use the fact that
i^2=-1


=i(-1)^2+(1)/(i(-1)^(12))+i(-1)^(22)

Use the fact that
(-1)^(a)=1 when a is an even number


=i+(1)/(i)+i

Simplify


=i-i+i


=i

Let me know if you need any clarifications, thanks!

User Shoeb Mirza
by
5.0k points