How do you simplify i^(-43)+i^(-32) ?

Apr 11, 2017

${i}^{- 43} + {i}^{- 32} = i + 1$

Explanation:

Look at the first few non-negative powers of $i$:

${i}^{0} = 1$

${i}^{1} = i$

${i}^{2} = - 1$

${i}^{3} = - i$

${i}^{4} = 1$

Basically this pattern: $1 , i , - 1 , - i$ repeats every $4$ powers.

In terms of angles, multiplying by $i$ is an anticlockwise rotation of $\frac{\pi}{2}$ in the complex plane. So after $4$ rotations we are back facing the same way.

So in general we can write:

$\left\{\begin{matrix}{i}^{4 n} = 1 \\ {i}^{4 n + 1} = i \\ {i}^{4 n + 2} = - 1 \\ {i}^{4 n + 3} = - i\end{matrix}\right.$

which holds for any integer $n$.

Now:

$- 43 = - 44 + 1 = 4 \left(- 11\right) + 1$

$- 32 = - 32 + 0 = 4 \left(- 8\right) + 0$

So:

${i}^{- 43} + {i}^{- 32} = i + 1$