# Thread: Using De Moivre's Theorem

1. ## 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

2. ## 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)]

3. ## Re: Using De Moivre's Theorem

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

Also state De Moivre's Theorem.

4. ## Re: Using De Moivre's Theorem

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

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.

5. ## Re: Using De Moivre's Theorem

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)

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.
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*}.