1. ## Comlex numbers

Hi.

I wonder if anyone know how to solve this math problem?

z^9 = 1 + i * square root of 3 , answer on the form a + bi

Should I solve it like this:
z^9= 2^9(cos (pi/3 *9) + isin (pi/3 *9)) = -512

or like this:
r^9(cos 9u + isin9u) = 2(cos 60 + isin 60)
n=0 zo=2^(1/9)(cos 60/9 + isin 60/9) = 2^(1/9) *(cos( 60/9) + isin (60/9))
n=1
.
.
.
n=9

Which is the right way to solve this problem?
Tank you.

2. Originally Posted by fgg
Hi.

I wonder if anyone know how to solve this math problem?

z^9 = 1 + i * square root of 3 , answer on the form a + bi

Should I solve it like this:
z^9= 2^9(cos (pi/3 *9) + isin (pi/3 *9)) = -512

or like this:
r^9(cos 9u + isin9u) = 2(cos 60 + isin 60)
n=0 zo=2^(1/9)(cos 60/9 + isin 60/9) = 2^(1/9) *(cos( 60/9) + isin (60/9))
n=1
.
.
.
n=9

Which is the right way to solve this problem?
Tank you.
Does giving one root solve a problem with 9?

$\displaystyle \displaystyle w_k=r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]$

$\displaystyle k\in\mathbb{Z}, \ 0\leq k\leq 8, \ \theta=\frac{\pi}{3}, \ n=9, \ r=2$

3. $\displaystyle \displaystyle z^9 = 1 + i\sqrt{3}$.

$\displaystyle \displaystyle |z^9| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

$\displaystyle \displaystyle \arg{(z^9)} = \arctan{\left(\frac{\sqrt{3}}{1}\right)} = \arctan{(\sqrt{3})} = \frac{\pi}{3}$.

So $\displaystyle \displaystyle z^9 = 2e^{i\frac{\pi}{3}}$

$\displaystyle \displaystyle z = \left(2e^{i\frac{\pi}{3}}\right)^{\frac{1}{9}}$

$\displaystyle \displaystyle z = \sqrt[9]{2}\,e^{i\frac{\pi}{27}}$.

This is the first of the nine square roots. The rest are evenly spaced around a circle, so have the same magnitude and are all separated by an angle of $\displaystyle \displaystyle \frac{2\pi}{9}$.

So the roots are $\displaystyle \displaystyle \sqrt[9]{2}\,e^{i\frac{\pi}{27}}, \sqrt[9]{2}\,e^{i\left(\frac{\pi}{27} + \frac{2\pi}{9}\right)}, \sqrt[9]{2}\,e^{i\left(\frac{\pi}{27} + \frac{4\pi}{9}\right)}, \dots$

Evaluate the rest of the roots, then convert them back to Cartesians.

4. Thank you very much for your help!