# How do you simplify i^-33?

May 27, 2015

Algebraically you could do this like this: ${i}^{- 33} = \frac{1}{{i}^{33}} = \frac{1}{{i}^{32} \cdot i} = \frac{1}{{\left({i}^{2}\right)}^{16} \cdot i} = \frac{1}{{\left(- 1\right)}^{16} \cdot i} =$
$= \frac{1}{1 \cdot i} = \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{{i}^{2}} = \frac{i}{- 1} = - i$

If you have to use the trigonometric form it goes like this:
${i}^{- 33} = {\left[1 \cdot \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)\right]}^{- 33} =$
$= {1}^{- 33} \cdot \left(\cos \left(\frac{- 33 \pi}{2}\right) + i \sin \left(\frac{- 33 \pi}{2}\right)\right) =$
$= \cos \left(\frac{33 \pi}{2}\right) - i \sin \left(\frac{33 \pi}{2}\right) =$
$= \cos \left(8 \cdot 2 \pi + \frac{\pi}{2}\right) - i \sin \left(8 \cdot 2 \pi + \frac{\pi}{2}\right) =$
$= \cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right) =$
$= 0 - i \cdot 1 = - i$
What I used here is the De Moivre's formula and some basic properties of $\sin$ and $\cos$.