pka
Elite Member
- Joined
- Jan 29, 2005
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There are six sixth roots of −12+5i see here.
See that there are six of them. Each of them is on a circle centered at the origin of radius ∣−12+5i∣1/6=613
However, the salient feature is the fact the the are evenly spaced on the circle and the central angle between any to adjacent is 62π=3π. see here
Here is a step by step outline: If N>1 is a positive integer and z=x+yi is a complex number in which x⋅y=0 (if that case fails there are special considerations).
Because all of this necessitates z 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 \(\displaystyle -6-8i\).

However, the salient feature is the fact the the are evenly spaced on the circle and the central angle between any to adjacent is 62π=3π. see here
Here is a step by step outline: If N>1 is a positive integer and z=x+yi is a complex number in which x⋅y=0 (if that case fails there are special considerations).
Because all of this necessitates z 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 \(\displaystyle -6-8i\).