Question

$i^5+i^{-25}+i^{45}$$i^5+i^{-25}+i^{45}$

Answer 1

$i^5+i^{-25}+i^{45}$

Rewrite the term $i^{-25}$

$=i^5+\dfrac{1}{i^{25}}+i^{45}$

Expand each term so we have

$=i(i^2)^2+\dfrac{1}{i(i^2)^{12}}+i(i^2)^{22}$

Use the fact that $i^2=-1$

$=i(-1)^2+\dfrac{1}{i(-1)^{12}}+i(-1)^{22}$

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

$=i+\dfrac{1}{i}+i$

Simplify

$=i-i+i$

$=i$

Let me know if you need any clarifications, thanks!

Answer 2

Answer:

1/i +2i

Step-by-step 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