Question

Find the value of
$\frac{i^6+i^7+i^8+i^9}{i^2+i^3}$

Answer 1

Now,

$\frac{i^6+i^7+i^8+i^9}{i^2+i^3}$

$=\frac{i^4.i^2+i^4.i^3+i^4.i^4+i^8.i}{i^2+i^3}$

$=\frac{-1-i+1+i}{-1-i}$ [ Since $i=\sqrt{-1}\Rightarrow i^2=-1$ and $i^4=1$]

$=0$.