High roots of negative numbers.

I was working a simple equation with the sine function, sin(x), and I converted the function to (e^(i*x))-(e^(i*-x))... later on I got to the 10th root of negative 1... so I just converted that to i because (i^10 = -1)... then the equation fell apart and nothing was working after that... does the 10th root of negative 1 have more answers than just "i"? If so, how do I find these roots?

Re: High roots of negative numbers.

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**orange gold** does the 10th root of negative 1 have more answers than just "i"? If so, how do I find these roots?

There are ten tenth roots of -1.

Let $\displaystyle \sigma=\exp\left(\frac{i\pi}{10}\right)~\&~\xi= \exp \left(\frac{i\pi}{5}\right)$.

The ten roots are: $\displaystyle \sigma\cdot\xi^k,~k=0,1,\cdots 9$

Re: High roots of negative numbers.

I'm a little unfamiliar with the notation you used. Would the values be complex numbers in the form x+iy? If so how can I find the x and y values? Would cosine and sine be related (assuming this based of your use of pi/n, resembles radian form to me)

Re: High roots of negative numbers.

Quote:

Originally Posted by

**orange gold** I'm a little unfamiliar with the notation you used. Would the values be complex numbers in the form x+iy? If so how can I find the x and y values? Would cosine and sine be related (assuming this based of your use of pi/n, resembles radian form to me)

The notation is from complex analysis.

It is simple: $\displaystyle \exp \left(i\theta\right)=\cos(\theta)+i\sin(\theta)$.

Thus $\displaystyle \left[\exp(i\theta)\right]^k=\cos(k\theta)+i\sin(k\theta)$.

It is easy notation for DeMoivre's Theorem.