# Thread: Find all complex roots

1. ## Find all complex roots

Hi,

I need to find all roots to the equation $\displaystyle z^4=-16i$

This is what i tried:

$\displaystyle |-16|^{1/4}=2$

And the angle of -16i is 270 degrees or $\displaystyle \frac{3\pi}{2}$

All the roots are on the form $\displaystyle |z|^{1/n}\cdot cos\frac{\theta}{n}+isin\frac{\theta}{n}$

So the first root should be $\displaystyle 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi gg)$

But it's just not correct.

2. Originally Posted by Jones
Hi,

I need to find all roots to the equation $\displaystyle z^4=-16i$

This is what i tried:

$\displaystyle |-16|^{1/4}=2$

And the angle of -16i is 270 degrees or $\displaystyle \frac{3\pi}{2}$

All the roots are on the form $\displaystyle |z|^{1/n}\cdot cos\frac{\theta}{n}+isin\frac{\theta}{n}$

So the first root should be $\displaystyle 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi gg)$

But it's just not correct.
Why do you say that? It is one of the roots.

3. Are you sure?

$\displaystyle 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi gg)$

gives $\displaystyle -2i, (-2i)^4 =16$

4. Originally Posted by Jones
I need to find all roots to the equation $\displaystyle z^4=-16i$
Here is the standard way to do all of these ‘root’ problems.
$\displaystyle -16i=16\exp\left(\frac{-\pi i}{2}\right)$.
Let $\displaystyle \omega =2\exp\left(\frac{-\pi i}{8}\right)$
Let $\displaystyle \xi =\exp\left(\frac{\pi i}{2}\right)$

Now it is really simple to list the four roots: $\displaystyle \omega \cdot \xi^k,~k=0,1,2,3$

5. Originally Posted by Jones
Are you sure?

$\displaystyle 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi gg)$

gives $\displaystyle -2i, \text{{\color{red} Mr F says: This is wrong. }} (-2i)^4 =16$
Of course I am sure.

6. An alternative method is to remember that if you're taking the 4th root, there will be 4 roots, and they are all evenly spaced around a circle.

7. Originally Posted by Jones
Are you sure?

$\displaystyle 2\bigg(cos(\frac{3\pi}{8})+isin(\frac{3\pi}{8})\bi gg)$

gives $\displaystyle -2i, (-2i)^4 =16$
No, it doesn't! $\displaystyle cos(\frac{3\pi}{8})= 0.38268$, not 0. cosine is 0 only at odd multiples of $\displaystyle \frac{\pi}{2}$.