Hi,

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

a. X^(3) - 1 = 10

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

Thanks in advance,

Lalit Chugh

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- Jun 19th 2006, 11:11 AMlalitchughDe'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

Thanks in advance,

Lalit Chugh - Jun 19th 2006, 01:34 PMThePerfectHackerQuote:

Originally Posted by**lalitchugh**

$\displaystyle x^3=11$

Express in polar form thus,

$\displaystyle x^3=11(\cos 2\pi k+i\sin 2\pi k)$

Thus,

$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3} \right)$ for $\displaystyle k=0,1,2$

Thus,

$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 0}{3}+i\sin \frac{2\pi \cdot}{3} \right)=11^{1/3}$

$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 1}{3}+i\sin \frac{2\pi \cdot 1}{3} \right)=11^{1/3}\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$

$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 2}{3}+i\sin \frac{2\pi \cdot 2}{3} \right)=11^{1/3}\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$ - Jun 19th 2006, 05:40 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**

- Jun 19th 2006, 05:59 PMThePerfectHackerQuote:

Originally Posted by**malaygoel**

$\displaystyle x^n=1$

We can then show that if,

$\displaystyle \zeta\in \left\{ \left{ \cos\frac{2\pi k}{n}+i\sin \frac{2\pi k}{n} \right| 0\leq k\leq n-1\right\}$

Then,

$\displaystyle \zeta^n=1$

Furthermore, all elements are distinct.

Thus, by the pigeonhole principle we have exactly $\displaystyle n$ elements in this set. But since a polynomial can have a most $\displaystyle n$ solution we see that all of the solution of, $\displaystyle x^n=1$ are found in this set.

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

Yes you can take other values for $\displaystyle k$, 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. - Jun 19th 2006, 09:33 PMlalitchughProblem 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 - Jun 21st 2006, 05:22 AMThePerfectHacker
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.