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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- December 15th 2011, 08:47 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 - December 15th 2011, 09:11 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)] - December 15th 2011, 09:12 AMPlatoRe: Using De Moivre's Theorem
- December 15th 2011, 05:06 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. - December 15th 2011, 05:15 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 to the angle of the first.

Note, you should really only list those angles that are within the range .