# How to get 4 roots for z^4 +16 =0?

1. Jun 18, 2013

### dla

1. The problem statement, all variables and given/known data
Solve for z^4 +16=0

2. Relevant equations

3. The attempt at a solution
What I first did was square rooted both sides to get z^2 = ±4i, but I don't how to continue from there. I'm guessing we have to find the roots from z^2=4i and then from z^2=-4i separately any help will be much appreciated!

2. Jun 18, 2013

### janhaa

try this first;

$$z^4+16=0$$

$$(z^2+4i)(z^2-4i)=0$$

3. Jun 18, 2013

### CAF123

You could try rewriting the RHS of $z^4 = -16$ in Euler form, which may make the problem less fiddly.

4. Jun 19, 2013

### HallsofIvy

In fact, in polar form, i= e^{i\pi/2} so $\sqrt{i}= e^{i\pi/4}= \sqrt{2}/2+ i\sqrt{2}/2$ and $e^{-i\pi/4}= \sqrt{2}/2- i\sqrt{2}{2}$

If $z^2= -4i$ then $z= \pm i(2)\sqrt{i}$.