pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,988
There are six sixth roots of see here.
See that there are six of them. Each of them is on a circle centered at the origin of radius
However, the salient feature is the fact the the are evenly spaced on the circle and the central angle between any to adjacent is . see here
Here is a step by step outline: If is a positive integer and is a complex number in which (if that case fails there are special considerations).
Because all of this necessitates being in polar form, I digress into this:
\(\displaystyle \arg(z)=\arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\dfrac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\dfrac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\dfrac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \) _______keep in mind that \(\displaystyle x\cdot y\ne 0 \text{ implies }x\ne 0~\&~y\ne 0\).
Return to the job at hand. If \(\displaystyle z=|z|\exp(\theta i)\) where \(\displaystyle \theta=\arg(z)\)
Define \(\displaystyle \zeta={\large\sqrt[N]{|z|}}\exp\left(\frac{\theta i}{N}\right)\) _____Do you see that \(\displaystyle \zeta^N=z~?\) I certainly hope you do !
Now we need a "rotator " to distribute the roots of \(\displaystyle z\) equality spaced around a circle. \(\displaystyle \rho=\exp\left(\frac{2\pi i}{N}\right)\)
The set \(\displaystyle \left\{\zeta\cdot\rho^k|k=0,1,\cdots N-1 \right\}\) is the collection of the \(\displaystyle N^{th}\) roots of \(\displaystyle z\).
Here is a practice question: List the seven seventh roots of \(\displaystyle -6-8i\).
See that there are six of them. Each of them is on a circle centered at the origin of radius However, the salient feature is the fact the the are evenly spaced on the circle and the central angle between any to adjacent is . see here
Here is a step by step outline: If is a positive integer and is a complex number in which (if that case fails there are special considerations).
Because all of this necessitates being in polar form, I digress into this:
\(\displaystyle \arg(z)=\arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\dfrac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\dfrac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\dfrac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \) _______keep in mind that \(\displaystyle x\cdot y\ne 0 \text{ implies }x\ne 0~\&~y\ne 0\).
Return to the job at hand. If \(\displaystyle z=|z|\exp(\theta i)\) where \(\displaystyle \theta=\arg(z)\)
Define \(\displaystyle \zeta={\large\sqrt[N]{|z|}}\exp\left(\frac{\theta i}{N}\right)\) _____Do you see that \(\displaystyle \zeta^N=z~?\) I certainly hope you do !
Now we need a "rotator " to distribute the roots of \(\displaystyle z\) equality spaced around a circle. \(\displaystyle \rho=\exp\left(\frac{2\pi i}{N}\right)\)
The set \(\displaystyle \left\{\zeta\cdot\rho^k|k=0,1,\cdots N-1 \right\}\) is the collection of the \(\displaystyle N^{th}\) roots of \(\displaystyle z\).
Here is a practice question: List the seven seventh roots of \(\displaystyle -6-8i\).