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- Jan 15th 2008, 12:54 PMyellow4321quick complex number question
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There are three possible values of z.

$\displaystyle z = 2e^{\frac{{\pi i}}{6}} \,,\,2e^{\frac{{ - \pi i}}{2}} \,\mbox{or} \,2e^{\frac{{5\pi i}}{6}} $ - Jan 15th 2008, 01:23 PMThePerfectHacker
Let $\displaystyle \zeta = e^{2\pi i/3}$.

And $\displaystyle \sqrt[3]{8i} = |8i|^{1/3} e^{i\arg(8i)/3} = 2 e^{i\pi/6} = 2 \cos \frac{\pi}{6} + 2 i\sin \frac{\pi}{6} = \sqrt{3} + i$.

Thus, the solution to $\displaystyle z^3 = 8i$ are: $\displaystyle (\sqrt{3}+i), (\sqrt{3}+i)\zeta, (\sqrt{3}+i)\zeta^2$. - Jan 15th 2008, 01:39 PMyellow4321
ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?

- Jan 15th 2008, 01:51 PMThePerfectHacker
**Theorem:**The polynomial $\displaystyle x^n - 1$ has $\displaystyle n$ roots over $\displaystyle \mathbb{C}$. Furthermore, if $\displaystyle \zeta = e^{2\pi i/n}$ then $\displaystyle \zeta, \zeta^2, ... ,\zeta^n $ are these solutions.

**Proof:**We use a fact that the polynomial $\displaystyle f(x) = x^n - 1$ can have*at most*$\displaystyle n$ zeros. If we let $\displaystyle \zeta = e^{2\pi i/n}$ then it is evident that $\displaystyle \zeta^n = e^{2\pi i} = 1$ and so $\displaystyle f(\zeta) = 0$. In fact if $\displaystyle 1\leq k \leq n$ then $\displaystyle (\zeta^k)^n = (\zeta^n)^k = 1^k = 1$. Thus, $\displaystyle f(\zeta^k) = 0$. Now we show that all $\displaystyle \zeta,...,\zeta^n$ are distinct. Suppose that $\displaystyle \zeta^a = \zeta^b$ then $\displaystyle \zeta^{(a-b)} = 1$ and so $\displaystyle \cos \frac{2\pi (a-b)}{n} + i\sin \frac{2\pi (a-b)}{n} = 1 + 0i$ that means $\displaystyle \frac{2\pi (a-b)}{n} = 0$ because $\displaystyle 1\leq a,b\leq n$ so $\displaystyle a=b$. Thus, the set $\displaystyle \{ \zeta , ... ,\zeta^n =1\}$ is a complete set of zeros for $\displaystyle f(x)$.

**Theorem:**If $\displaystyle a\not = 0$ is a complex number then $\displaystyle f(x) = x^n - a$ has exactly $\displaystyle n$ zeros and those are $\displaystyle a^{1/n}\zeta, a^{1/n}\zeta^2, ... ,a^{1/n}\zeta^n$.

**Proof:**First the polynomial $\displaystyle f(x)$ can have at most $\displaystyle n$ zeros so if we can show that these are all zeros (and that they are distinct) it will complete the proof. First $\displaystyle a^{1/n}$ is the principle $\displaystyle n$-root defined at $\displaystyle |a|^{1/n} e^{i\arg(a)/n}$, and so $\displaystyle \left( a^{1/n} \right)^n = 1$. Now if $\displaystyle 1\leq k \leq n$ it means $\displaystyle \left( a^{1/n} \zeta^k \right)^n = \left( a^{1/n}\right)^n \left( \zeta^k \right)^n = 1$. Which means $\displaystyle a^{1/n} \zeta^k$ are all zeros. Finally to show there are all distinct it is easy. Suppose $\displaystyle a^{1/n} \zeta^r = a^{1/n} \zeta^s$ where $\displaystyle 1\leq r,s \leq n$ then since $\displaystyle a\not = 0$ it means $\displaystyle \zeta^r = \zeta^s \implies r=s$, as in the previous proof. And so this means that all these zeros are distinct. - Jan 15th 2008, 02:27 PMPlato
- Jan 15th 2008, 02:28 PMyellow4321
ok i sort of understand, and now know the fact that you have n-1 solutions if n is the power so solve for k=0,1,2..n-1 but could you show me how to get to the 5pi/6 and -pi/2 why arent i just added pi to pi/6 to get 7pi/6 or something.

arrrr i see so if i worked in degrees i could just do 360/3 which is 120 then add and minus this 30, then il get the radian form.

you guys are my heros