# Thread: Easy imaginary numbers problem?

1. ## Easy imaginary numbers problem?

Simplify i^-19?

Thanks!

2. ## Re: Easy imaginary numbers problem?

Originally Posted by woahitzme
Simplify i^-19?

Thanks!
\displaystyle \begin{align*} i^{-19} &= \left(e^{\frac{\pi}{2}i}\right)^{-19} \\ &= e^{-\frac{19\pi}{2}i} \\ &= e^{\left(-4\pi - \frac{3\pi}{2}\right)i} \\ &= e^{-\frac{3\pi}{2}i} \\ &= e^{\frac{\pi}{2}i} \\ &= i \end{align*}

3. ## Re: Easy imaginary numbers problem?

Originally Posted by Prove It
\displaystyle \begin{align*} i^{-19} &= \left(e^{\frac{\pi}{2}i}\right)^{-19} \\ &= e^{-\frac{19\pi}{2}i} \\ &= e^{\left(-4\pi - \frac{3\pi}{2}\right)i} \\ &= e^{-\frac{3\pi}{2}i} \\ &= e^{\frac{\pi}{2}i} \\ &= i \end{align*}

Umm...

$i^{20} = 1$

$i^{-19} = i^{-19}i^{20} = i$

4. ## Re: Easy imaginary numbers problem?

Originally Posted by richard1234
Umm...

$i^{20} = 1$

$i^{-19} = i^{-19}i^{20} = i$
I preferred to show a method that allows you to exponentiate ANY complex number

5. ## Re: Easy imaginary numbers problem?

Good point...