Using De Moivre's Theorem find all the solutions (z) to he equation:

Z^6-1=jroot3

j being the imaginary number

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- Dec 15th 2011, 07:47 AMcalmathUsing De Moivre's Theorem
Using De Moivre's Theorem find all the solutions (z) to he equation:

Z^6-1=jroot3

j being the imaginary number - Dec 15th 2011, 08:11 AMTheChazRe: Using De Moivre's Theorem
So z^6 = 1 + i*sqrt(3)

Can you write the right side in polar/trig form?

z^6 = 2[cos(60) + i*sin(60)] - Dec 15th 2011, 08:12 AMPlatoRe: Using De Moivre's Theorem
- Dec 15th 2011, 04:06 PMcalmathRe: Using De Moivre's Theorem
De moivre's theorem; Z^n=r^n[cos(n theta) + i sin (n theta)]

The answers that im getting feels like not correct.

Polar form:

r=sqrt(10)

Theta=1.25radians

z=10^1/12(cos(2npi+1.25/6)+i sin (2npi+1.25/6)) I used this to get values for each z power, and this is what i calculated;

z= 1.4 rads, z1= 1.5rads, z2=0.1rads, z3=1.5rads, z4=-1.5rads, z5=-0.1rads

I believe I should be getting hexagon shape when i plot these numbers on an argand diagram but i wont be able to as some of my calculation are wrong since all sides of hexagon should be equal. Could you pls help me find where im making the mistake. - Dec 15th 2011, 04:15 PMProve ItRe: Using De Moivre's Theorem
When you say "hexagon shape", I assume you mean that all six roots should be evenly spaced about a circle.

That means that if you find one of the sixth roots, you can find the others by adding $\displaystyle \displaystyle \begin{align*} \frac{2\pi}{6} = \frac{\pi}{3}\end{align*}$ to the angle of the first.

Note, you should really only list those angles that are within the range $\displaystyle \displaystyle \begin{align*} -\pi < \theta \leq \pi \end{align*}$.