1. quick complex number question

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2. There are three possible values of z.
$z = 2e^{\frac{{\pi i}}{6}} \,,\,2e^{\frac{{ - \pi i}}{2}} \,\mbox{or} \,2e^{\frac{{5\pi i}}{6}}$

3. Originally Posted by yellow4321
if z^3=8i
i know very little about complex numbers.
i used euler to get rt(3) +i
but im having doubts would the answer be +/- 2i ? i thought that too easy.
Let $\zeta = e^{2\pi i/3}$.
And $\sqrt[3]{8i} = |8i|^{1/3} e^{i\arg(8i)/3} = 2 e^{i\pi/6} = 2 \cos \frac{\pi}{6} + 2 i\sin \frac{\pi}{6} = \sqrt{3} + i$.

Thus, the solution to $z^3 = 8i$ are: $(\sqrt{3}+i), (\sqrt{3}+i)\zeta, (\sqrt{3}+i)\zeta^2$.

4. ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?

5. Originally Posted by yellow4321
ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?
Theorem: The polynomial $x^n - 1$ has $n$ roots over $\mathbb{C}$. Furthermore, if $\zeta = e^{2\pi i/n}$ then $\zeta, \zeta^2, ... ,\zeta^n$ are these solutions.

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

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

Proof: First the polynomial $f(x)$ can have at most $n$ zeros so if we can show that these are all zeros (and that they are distinct) it will complete the proof. First $a^{1/n}$ is the principle $n$-root defined at $|a|^{1/n} e^{i\arg(a)/n}$, and so $\left( a^{1/n} \right)^n = 1$. Now if $1\leq k \leq n$ it means $\left( a^{1/n} \zeta^k \right)^n = \left( a^{1/n}\right)^n \left( \zeta^k \right)^n = 1$. Which means $a^{1/n} \zeta^k$ are all zeros. Finally to show there are all distinct it is easy. Suppose $a^{1/n} \zeta^r = a^{1/n} \zeta^s$ where $1\leq r,s \leq n$ then since $a\not = 0$ it means $\zeta^r = \zeta^s \implies r=s$, as in the previous proof. And so this means that all these zeros are distinct.

6. Originally Posted by yellow4321
ok thanks i understand how to get 2e^i*(pi/6) but the over two solutions i do not understand how to achieve?
The three cube roots are located on a circle with center at the origin and radius 2.
On that circle the roots are equally ‘spread’ out.
Because you know one is at ${\frac{\pi}{6}}$, see how the other two are spread out.

7. 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