Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $None.$

Answer 1

The correct option is A $0$

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

$=\frac{(i^2{)}^3-(i^2{)}^{3}i+(i^4{)}^2+(i^4{)}^{2}i}{-1-i}$

We know that, $i^2=-1$ and $i^4=1$

$=\frac{-1-i+1+i}{-1-i}$

$Z=0$

$\therefore \frac{i^6+i^7+i^8+i^9}{i^2+i^3}=0$