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

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Homework Statement


Solve for z^4 +16=0

Homework Equations


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!
 
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try this first;

z^4+16=0

(z^2+4i)(z^2-4i)=0
 
You could try rewriting the RHS of ##z^4 = -16## in Euler form, which may make the problem less fiddly.
 
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}.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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