14.6k views
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
7.2k 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
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories