# Thread: Problem reg De'Moivre's theorem

1. ## De'Moivre's theorem

Hi,

Use De’Moivre’s theorem to solve the following equations:

a. X^(3) - 1 = 10

b. (x + 1)^(5) + x^(2) = 0

Lalit Chugh

2. Originally Posted by lalitchugh
Hi,

Use De’Moivre’s theorem to solve the following equations:

a. X^(3) - 1 = 10
Thus,
$x^3=11$
Express in polar form thus,
$x^3=11(\cos 2\pi k+i\sin 2\pi k)$
Thus,
$x=11^{1/3}\left( \cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3} \right)$ for $k=0,1,2$
Thus,
$x=11^{1/3}\left( \cos \frac{2\pi \cdot 0}{3}+i\sin \frac{2\pi \cdot}{3} \right)=11^{1/3}$
$x=11^{1/3}\left( \cos \frac{2\pi \cdot 1}{3}+i\sin \frac{2\pi \cdot 1}{3} \right)=11^{1/3}\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$
$x=11^{1/3}\left( \cos \frac{2\pi \cdot 2}{3}+i\sin \frac{2\pi \cdot 2}{3} \right)=11^{1/3}\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$

3. Originally Posted by ThePerfectHacker
Thus,
$x^3=11$
Express in polar form thus,
$x^3=11(\cos 2\pi k+i\sin 2\pi k)$
Thus,
$x=11^{1/3}\left( \cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3} \right)$ for $k=0,1,2$
I think we could take any value of k, why only 0,1 and 2, why not 3,4 and 5?

4. Originally Posted by malaygoel
I think we could take any value of k, why only 0,1 and 2, why not 3,4 and 5?
Because given,
$x^n=1$
We can then show that if,
$\zeta\in \left\{ \left{ \cos\frac{2\pi k}{n}+i\sin \frac{2\pi k}{n} \right| 0\leq k\leq n-1\right\}$
Then,
$\zeta^n=1$
Furthermore, all elements are distinct.
Thus, by the pigeonhole principle we have exactly $n$ elements in this set. But since a polynomial can have a most $n$ solution we see that all of the solution of, $x^n=1$ are found in this set.

I like to think this is as elegant as it gets.

Yes you can take other values for $k$, but I like to follow this reasoning. Because, I am certain that all the solutions are found by starting at k=0 and going to k=n-1.

5. ## Problem reg De'Moivre's theorem

Hi,

Many thanks for the solution.

May i request you to provide solution for 2nd question on De'Moivre's thoerem as well.

I will be thankful to you. It will help me in undertstanding the steps.

Best Regards,
Lalit Chugh

6. latichugh do not beg!
-=USER WARNED=-
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Your question was not answered perhaps because it has nothing with de Moiver's theorem.