Question

Please anyone help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I would really appreciate it

Answer 1

If $i = \sqrt{-1}$, then $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. For larger integer powers of i, the cycle repeats:

$i^5 = i^4\cdot i^1 = i \\\\ i^6 = i^4\cdot i^2 = -1 \\\\ i^7 = i^4\cdot i^3 = -i \\\\ i^8 = i^4\cdot i^4 = 1$

and so on.

Then

(1)

$i^{8n} = i^{4\cdot2n} = \left(i^4\right)^{2n} = 1^{2n} = 1$

(2)

$i^{4n+42} = i^{4n+40+2} = i^{4n+40}\cdot i^2 = \left(i^4\right)^{n+10}\cdot i^2 = 1^{n+10}\cdot i^2 = -1$

(3)

$i^{12n+3} = i^{12n}\cdot i^3 = \left(i^4\right)^{3n} \cdot i^3 = 1^{3n}\cdot i^3 = -i$

(4)

$i^{8n-3} = i^{8n}\cdot i^{-3} = \left(i^4\right)^{2n}\cdot \dfrac1{i^3} = \dfrac{1^{2n}}{i^3} = \dfrac1{i^3} = -\dfrac1i = i$