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
Thus,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,
$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 0}{3}+i\sin \frac{2\pi \cdot}{3} \right)=11^{1/3}$
$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 1}{3}+i\sin \frac{2\pi \cdot 1}{3} \right)=11^{1/3}\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$
$\displaystyle x=11^{1/3}\left( \cos \frac{2\pi \cdot 2}{3}+i\sin \frac{2\pi \cdot 2}{3} \right)=11^{1/3}\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$
Because given,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.