# Using De Moivre's Theorem

• Dec 15th 2011, 07:47 AM
calmath
Using 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 AM
TheChaz
Re: 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 AM
Plato
Re: Using De Moivre's Theorem
Quote:

Originally Posted by calmath
Using De Moivre's Theorem find all the solutions (z) to he equation:
Z^6-1=jroot3

Here is a start: $1+\sqrt{3}i=2[\cos\left(\frac{\pi}{3}\right)+i~\sin\left(\frac{ \pi}{3}\right)].$

Also state De Moivre's Theorem.
• Dec 15th 2011, 04:06 PM
calmath
Re: Using De Moivre's Theorem
Quote:

Originally Posted by Plato
Here is a start: $1+\sqrt{3}i=2[\cos\left(\frac{\pi}{3}\right)+i~\sin\left(\frac{ \pi}{3}\right)].$

Also state 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)

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;

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 PM
Prove It
Re: Using De Moivre's Theorem
Quote:

Originally Posted by calmath
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)
That means that if you find one of the sixth roots, you can find the others by adding \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 \begin{align*} -\pi < \theta \leq \pi \end{align*}.